Patterns in Pythagoras
3B. Even More Patterns in Right Triangles
andPythagorean Triples
David W. Hansen
© 2008
4. Unit and Constant Difference Triples
A unit primitive Pythagorean triple? Just what is that? Well, as the name suggests, it is a primitive Pythagorean triple (a, b, c), where the difference between a and b is the unit 1. (3,4,5) is a unit Pythagorean triple because b - a = 4 - 3 = 1. (21,20,29) is a unit triple because a - b = 21 - 20 = 1. Unit triples can be regular (a < b) as in (3,4,5) or inverted (a > b) as in (21,20,29). If |a – b| = 1, then we call the primitive Pythagorean triple (a, b, c) a unit triple. Are there unit triples other than the two just mentioned? Let’s see.
A. If b = a+1, then, the triple (a, a+1, c) is a regular unit triple, and a2 + (a+1)2 = c2. Solving for a, we get _______
a2 + a2 + 2a + 1 = c2, 2a2 + 2a + (1 - c2) = 0, and a = ( - 1 ± Ö 2c2 -1 ) / 2.
B. If b = a -1, then, the triple (a, a -1, c) is an inverted unit triple, and a2 + (a-1)2 = c2. Solving for a, we
get _______
a2 + a2 - 2a + 1 = c2, 2a2 - 2a + (1 - c2) = 0, and a = ( 1 ± Ö 2c2 -1 ) / 2.
Now a must be an integer, so in both cases A and B above, 2c2 –1 must be a perfect square in order for its square root to be an integer. Furthermore, since c is odd, then c2 is odd, 2c2 is even, 2c2 – 1 is odd,
______ ______ ______
Ö 2c2 - 1 is odd, 1 ± Ö 2c2 -1 is even and thus divisible by 2. So, a = ( 1 ± Ö 2c2 -1 ) / 2 is an integer.
Thus, to find all unit triples, we simply need to find all those values of c for which 2c2 – 1 is a perfect square.
Let’s set up a table of the permissible values of c written in ascending order, together with their corresponding
values of 2c2 - 1, and if these are perfect squares, their square roots, the values of a, b, and c, and the values of m and n.
Table 3
______
c 2c2 -1 Ö 2c2 - 1 a b c m n
-------------------------------------------------------------------------------------------
5 49 7 3 4 5 2 1
13 337 -
17 577 -
25 1249 -
29 1681 41 21 20 29 5 2
37 2737 -
41 3361 -
53 5617 -
. . . . . . . . . . . .
157 49297 -
169 57121 239 119 120 169 12 5
173 59857 -
181 65521 -
---------------------------------------------------------------------------------------------
Now, as seen in Table 3, we have calculated the values of 2c2 - 1 for all permissible values of c up to 181 and found just one new unit triple. Are there others? If so, can we find a simpler and more pleasing way to find them with less computation involved? Let’s look for a pattern in Table 3 that might help us.
Writing down the three unit triples we have found and their corresponding values of m and n in the table below, we get
(a, b, c) m n
-------------------------------------------
(3, 4, 5) 2 1
(21, 20, 29) 5 2
(119, 120, 169) 12 5
What can we see? Well, the m value of the first unit triple (2) is the same as the n value of the second triple, and the m value of the second unit triple (5) is the same as the n value of the third unit triple! If this pattern continues, then the m value of the third unit triple (12) will be the same as the n value of the fourth unit triple, as shown below.
(a, b, c) m n
---------------------------------------------------
(3, 4, 5) 2 1
(21, 20, 29) 5 2
(119, 120, 169) 12 5
(?, ?, ?) m? 12
But what will the m value of the fourth unit triple be?
If we look closely at the values of m and n for the first two unit triples as shown below,
Unit triple 1 2 1
Unit triple 2 5 2
we see that the sum of the three bold-faced values , 2 + 1 + 2, equals the underlined value, 5. Thus, the m value of the second unit triple equals the combined sum of the m and n values of the first unit triple and the n value of the second unit triple.
Similarly, looking at the values of m and n for the second and third unit triples,
Unit triple 2 5 2
Unit triple 3 12 5
we see that the sum of the three bold-faced values, 5 + 2 + 5, equals the underlined value, 12. Thus, the m value of the third unit triple equals the sum of the m and n values of the second unit triple and the n value of the third unit triple.
Now if the pattern continues, and looking at the values of m and n for the third and fourth unit triples as
shown below, we surmise that the m value for the fourth unit triple will equal the combined sum of the m and n values of the third unit triple and the n value of the fourth unit triple; namely, 12 + 5 + 12 = 29.
Unit triple 3 12 5
Unit triple 4 m? 12
Thus, the m value for the fourth unit triple would be 29. Let’s check this.
Using m = 29, n = 12, will we obtain a unit Pythagorean triple? Let’s see! a = 292 - 122 = 697,
b = 2(29)(12) = 696, and c = 292 + 122 = 985, and we see that indeed (697, 696, 985) is the fourth unit triple, since a - b = 1. How VERY nice!
Adding this unit triple to our list of unit triples, we get
(a, b, c) m n
-----------------------------------------------
(3, 4, 5) 2 1
(21, 20, 29) 5 2
(119, 120, 169) 12 5
(697, 696, 985) 29 12
Notice that the first unit triple is regular, the second triple is inverted, the third one is regular, and the fourth one is inverted. Does this pattern continue for all unit triples? Are there more unit triples? Are there an infinite number of unit triples? We surely hope so!
Now what have we found so far? It seems that when we have a list of unit triples sorted in ascending order by their c values, then the m value of a given unit triple U equals the n value of the next unit triple, the combined sum of the m and n values of U and the n value of the next unit triple equals the m value of the next unit triple, and the unit triples alternate between being regular and inverted triples.
In general, in a list of unit triples sorted in ascending order by their c values, if U1 is a unit triple with
m = m1, n = n1, and U2 is the next unit triple with m = m2, n = n2, then the list looks like this:
(a, b, c) m n
----------------------------------------------
. . . . . . . . .
U1 m1 n1
U2 m2 n2
. . . . . . . . .
-----------------------------------------------
and we surmise that n2 = m1 and m2 = m1 + n1 + n2 , or, since n2 = m1, then m2 = 2m1 + n1, making the list now look like this:
(a, b, c) m n
------------------------------------------------------
. . . . . . . . .
U1 m1 n1
U2 2m1+n1 m1
. . . . . . . . .
----------------------------------------------
Let’s see if we can prove this conjecture. Accordingly, we let U1 be any regular unit triple (a1, b1, c1) in a list of unit triples sorted in ascending order by their c values, with m = m1, n = n1. Since U1 is a regular unit triple, we know that
b1 = a1 + 1,
or 2m1n1 = m12 – n12 + 1. (1)
and multiplying (1) by -1, we have - 2m1n1 = n12 – m12 – 1 (2)
Next, we let U2 be the next triple (a2, b2, c2) in the list with its m and n values given by m = 2m1 + n1 and n = m1. Will U2 be a unit triple for these values of m and n? To show that it is, we must show that either b2 = a2 + 1 or
b2 = a2 - 1.
Now, b2 = 2mn = 2(2m1 + n1)m1 = 4m12 + 2m1n1, (3)
and a2 - 1 = m2 - n2 - 1 = (2m1 + n1)2 - m12 - 1
= 4m12 + 4m1n1 + n12 - m12 - 1
= 4m12 + 4m1n1 + (n12 - m12 - 1)
= 4m12 + 4m1n1 + ( - 2m1n1), (substituting from (2) above)
= 4m12 + 2m1n1, which is (3) above.
Thus, b2 = a2 - 1, and U2 is a unit triple. Furthermore, since b2 < a2, it is inverted !
If U1 had been an inverted unit triple (b1 = a1 - 1), our proof would have shown that U2, the next triple in the list, would be a regular unit triple (b2 = a2 + 1). Thus, in a list of unit triples sorted in ascending order by their c values, the triples will alternate between regular and inverted triples, and since (3,4,5), the first unit triple, is regular, the list will begin with a regular unit triple.
Note that for any given unit triple, we can always find the next one, so there are an infinite number of unit triples! All of this gives us the following theorem.
Theorem 29
If U is any unit triple in a list of unit triples sorted in ascending order by their c values,
with m = m1, and n = n1, then the next unit triple in this list has m and n values given by
m = 2m1 + n1 and n = m1. If U is regular, the next triple will be inverted. If U is inverted,
the next triple will be regular. There are an infinite number of unit triples, half of which are
regular and half of which are inverted.
________________________________________________________________________
To use Theorem 29, we must be able to find the values of m and n for any given unit triple (a, b, c). Here’s how. Since c = m2 + n2 and a = m2 - n2, then
c + a = (m2 + n2) + (m2 - n2) = 2m2, or m2 = (c + a) /2.
and, c - a = (m2 + n2) - (m2 - n2) = 2n2, or n2 = (c - a) /2.
_______ ______
So, for a primitive Pythagorean triple (a, b, c), m = Ö (c+a)/2 and n = Ö (c-a)/2 . (4)
Example 1. Find the next unit triple after (697, 696, 985). To use Theorem 29, we must first find the values of m and n for (697, 696, 985). Using (4) above, we get
___________ __________
m = m1 = Ö (985+697)/2 = 29 and n = n1 = Ö (985-697)/2 = 12,
which confirm the values shown in the table on page 19. Now, using Theorem 29, we have for the next unit triple,
m = 2m1 + n1 = 2(29) + 12 = 70, and n = m1 = 29.
To be pictorial, we could have used the following tables (which are more fun):
29 12 29 12 29 12 29 12
→ → →
? ? ? 29 (29+12+29) 29 70 29
These values for m and n give us a = 702 – 292 = 4059, b = 2mn = 2(70)(29) = 4060, and c = 702 +292
= 5741, thus generating (4059, 4060, 5741), the next unit triple after (697, 696, 985). Note that this new unit triple is regular, since (697, 696, 985) was inverted.
Example 2. U = (137903, 137904, 195025) is a unit triple. For the unit triple just preceding U, a) determine if it is regular or inverted, and b) find this triple.
a) Since U is a regular unit triple, (a < b), then by Theorem 29, the unit triple preceding U must be inverted
(b > a).
b) We must first find m and n for U. Thus, using (4) above, we get for U,
________________ ________________
m = Ö(195025+137903)/2 = 408, and n = Ö(195025+137903)/2 = 169.
Now. to find the preceding unit triple, we must “reverse” the procedure in Theorem 29.
Pictorially, we can use the following tables to work backwards to find the m1 and n1 values for the preceding unit triple. As shown in the tables below, we put m = 408 and n = 169 in row 2 of Table A. Then, since m1 = n, we put 169 as the first entry in row 1, giving us Table B. Since the sum of both the entries in row 1 and the second entry in row 2 must equal the first entry in row 2, we subtract the sum of the first entry in row 1 and the second entry in row 2 to find the second entry in row 1 (Table C below). This gives us Table D below, which tells us m1 = 169, n1 = 70.
Table A Table B Table C Table D
? ? 169 ? 169 (408 -169 -169) 169 70
→ → →
408 169 408 169 408 169 408 169
Algebraically, using Theorem 29, we can find the m1 and n1 of the preceding unit triple by solving the equations m = 2m1 + n1 and n = m1 for m1 and n1. Substituting the values of m = 408 and n = 169 into these two equations, we get
408 = 2m1 + n1 (5)
and 169 = m1.
Thus, m1 = 169, and substituting this value of m1 into (5), we get 408 = 2(169) + n1, or n1 = 70.
Either pictorially or algebraically, m1 = 169 and n1 = 70, giving us a = 1692 - 702 = 23661, b = 2mn
= 2(169)(70) = 23660, and c = 1692 + 702 = 3 3461, thus generating (23661 23660, 33461), the inverted unit triple (23661 > 23660) which precedes (137903, 137904, 195025).
Adding the information and results found in examples 1 and 2 to the table of unit triples above, we get Table 4 below, which lists the first seven unit triples together with their values of m and n.
Table 4 (Unit Triples)
(a, b, c) m n
-----------------------------------------------------------
(3, 4, 5) 2 1
(21, 20, 29) 5 2
(119, 120, 169) 12 5
(697, 696, 985) 29 12
(4059, 4060, 5741) 70 29
(23661, 23660, 33461) 169 70
(137903, 137904, 195025) 408 169
Now, does there exist any triple in which the difference between a and b is 1 and the difference between b and c is also 1? If so, we shall call that triple a pure unit triple. Let’s try to find some.
For a Pythagorean triple (a, b, c), if a and b differ by 1, then b = a + 1, and if b and c differ by 1, then c = b + 1
= (a + 1) + 1 = a + 2. Then, substituting for b and c in a2 + b2 = c2,
we have
a2 + (a + 1)2 = (a + 2)2,
a2 + a2 + 2a + 1 = a2 + 4a + 4,
and simplifying, we get a2 – 2a – 3 = (a – 3)(a + 1) = 0.
Solving this last equation, we have a = 3 or - 1. Since a > 0, we must pick a = 3, giving us a = 3, b = 3 + 1 = 4, and c = 3 + 2 = 5. Thus, (3,4,5) is a pure unit triple, and, in fact, the only pure unit triple!
Constant Differences between a, b, and c
Now, we have just finished looking for and finding patterns in those primitive Pythagorean triples where the difference between the a and b values is always 1. Let’s now look at those triples where the difference between their c and b values, their c and a values, and their and b values is always a constant other than 1.
The difference between c and b Since c – b = (m2 + n2) – 2mn = (m – n)2, then c – b is always a perfect square. Furthermore, since c is always odd and b is always even, then
c – b is always an odd perfect square.
Let c – b = d2, where d is some given odd positive integer. Then, we have
c – b = (m – n)2 = d2, or m – n = d.
Thus, m = n + d, and substituting this into the generating formulas for a, b, and c, we get
a = m2 – n2 = (n + d)2 – n2 = 2nd + d2
b = 2mn = 2(n + d)n = 2n(n + d)
c = m2 + n2 = (n + d)2 + n2
Now, n must not be a multiple of d, for if n = fd for some integer f, then substituting this into the above equations, we get
a = 2d(fd) + d2 = d2(2f + 1), b = 2fd(fd + d) = d2(2f2 + 2f),
and c = (fd + d)2 + (fd)2 = f2d2 + 2fd2 + d2 + f2d2 = d2(2f2 + 2f + 1).
Thus, a, b, and c will all have a common factor of d2, and (a, b, c) will not be a primitive Pythagorean triple
This gives us
Theorem 30 (The difference between c and b)
If (a, b, c) is a primitive Pythagorean triple, then c – b is always an odd perfect square, say d2.
All primitive Pythagorean triples (a, b, c) for which c – b = d2, where d is an odd positive integer,
are given by a = 2dn + d2, b = 2n(n + d), and c = (n + d)2 + n2, where n is any positive
integer not a multiple of d
______________________________________________________________________________
Example 3. Find formulas for all primitive Pythagorean triples (a, b, c) for which the difference between c and b is 9. We have c – b = 9 = 32 = d2. Thus, d = 3, and from Theorem 30, we get
a = 2(3)n + 32 = 6n + 9, b = 2n(n + 3), c = (n + 3)2 + n2
for n a positive integer not a multiple of 3. Below is a table of these triples for n = 1, 2, 4, 5, 7, and 8. Note that n ≠ 3, 6, 9, . . . , which are multiples of 3.
n a b c c – b n a b c c – b
-------------------------------------------------------------------------------------------------------------------
1 15 8 17 17 – 8 = 9 5 39 80 89 89 – 80 = 9
2 21 20 29 29 – 20 = 9 7 51 140 149 149 – 100 = 9
4 33 56 65 65 – 56 = 9 8 57 176 185 185 – 176 = 9
Example 4. Are there any primitive Pythagorean triples (a, b, c) for which the difference between c and b is 13, 36, 49, or 123? Since 13, 36, and 123 are not odd squares, there are no primitive triples for these values. However, 49 = 72 is an odd square with d = 7. So, a = 2(7)n + 72 = 14n + 49, b = 2n(n + 7), and c = (n + 7)2 + n2, and the triples (14n + 49, 2n(n + 7), (n + 7)2 + n2) for n a positive integer not a multiple of 7, all have a difference of 49 between their b and c values. This is so because
c – b = (n + 7)2 + n2 – 2n(n + 7) = n2 + 14n + 49 + n2 – 2n2 – 14n = 49 √
The difference between c and a Since c – a = (m2 + n2) – (m2 – n2) = 2n2, then
c – a is always twice a perfect square.
Here is a table showing the possible values of c – a for n = 1, 2, 3, . . . , 15.
n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
------------------------------------------------------------------------------------------------------------------------------------
c – a 2 8 18 32 50 72 98 128 162 200 242 288 338 392 450
Now, m and n are of opposite parity (one even, one odd), so m – n is always odd; that is, m – n = 2k - 1
for some positive integer k. Solving this for m, we get m = n + 2k – 1 and substituting this for m in the generating formulas for a, b, and c, we have
a = m2 – n2 = (n + 2k – 1)2 – n2 (6)
b = 2mn = 2(n + 2k – 1) = 2n(n + 2k – 1) (7a)
c = m2 + n2 = (n + 2k – 1)2 + n2 (7b)
Now, a, b, and c must have no common factors. Let p be a prime factor of n. Then, n = pg for some positive integer g, and substituting this value of n into (7a), we see that
b = 2n(n + 2k – 1) = 2pg(n + 2k – 1) = p[2g(n + 2k – 1)],
which shows that b has a factor of p. Furthermore, substituting this value of n into (6), we get
a = (n + 2k – 1)2 – n2 = [pg + (2k – 1)]2 – (pg)2,
and looking at this last expression, we see that if 2k – 1 is a multiple of p (that is, 2k – 1 = ph for some positive integer h), then
a = [pg + (2k – 1)]2 – (pg)2 = [pg + ph]2 – (pg)2
= [p(g + h) ]2 – (pg)2 = p2(g + h)2 – p2g2
= p [ p(g + h2 – pg2 ], which shows that a has a factor of p.
Thus, 2k – 1 cannot be a multiple of a prime factor p of n, or else a and b will have a common factor of p, and
(a, b, c) will not be primitive.
Now, if 2k – 1 = ph for some integer h, then both p and h must be odd, since 2k – 1 is odd. Rewriting
2k – 1 = ph as 2k = ph + 1 = ph + p + 1 – p = p(h + 1) – (p – 1),
we get k = p[(h+1)/2] – (p – 1)/2.
Now, p and h are both odd, so (h+1)/2 is an integer, say w, as is (p – 1)/2 also. This gives us
k = pw – (p – 1)/2, for w = 1, 2, 3, . . .
Thus, if (a, b, c) is to be primitive, then k must not be equal to pw – (p – 1)/2, in (6), (7a), and (7b), giving us
Theorem 31 (The difference between c and a)
If (a, b, c) is a primitive Pythagorean triple, then c – a is always twice a perfect square, namely, 2n2,
and
a = (n + 2k – 1)2 – n2, b = 2n(n + 2k – 1), c = (n + 2k – 1)2 + n2,
where n and k are any positive integers with k ≠ pw – (p – 1)/2 for p an odd prime factor of n,
and w = 1, 2, 3, . . .
_______________________________________________________________________________
Example 5. Find formulas for all primitive Pythagorean triples (a, b, c) for which the difference between c and a is 98. Using Theorem 31, we know that c – a = 2n2 = 98. Thus, n2 = 98, n = 7, and p = 7. Then,
a = (2k + 6)2 – 49, b = 14(2k + 6), c = (2k + 6)2 + 49
for k ≠ 7w – 3, w = 1, 2, 3, . . . , that is, k ≠ 4, 11, 18, 25, . . .
Below is a table of these triples for k = 1, 2, 3, 5, 6, 7, 8, and 9.
k a b c c – a k a b c c – a
--------------------------------------------------------------------------------------------------------------
1 15 112 113 98 6 275 252 373 98
2 51 140 149 98 7 351 280 449 98
3 95 168 193 98 8 435 308 533 98
5 207 224 305 98 9 527 336 625 98
Example 6. Find formulas for all primitive Pythagorean triples (a, b, c) for which the difference between
c and a is 450. From Theorem 31, we know that c – a = 2n2 = 450, so, n2 = 225, and n = 15 = 3(5). Thus, n has two odd prime factors; namely, p = 3 and p = 5.
For p = 3, we get k ≠ pw – (p – 1)/2 = 3w – 1 , and for p = 5, we get k ≠ 5w – 2.
Thus, for n = 15, we have a = (2k + 14)2 – 225
b = 30(2k + 14)
c = (2k + 14)2 + 225
for k ≠ 3w – 1, w = 1, 2, 3, 4, . . . and k ≠ 5w – 2, w = 1, 2, 3, 4, . . .
that is, k ≠ 2, 5, 8, 11, 14, 17, . . . and k ≠ 3, 8, 13, 18, 23, . . .
Below is a table of these triples for k = 1, 4, 6, 7, 9, and 10. Note that k ≠ 2, 3, 5, 8, 11, 13, . . .
k a b c c – a k a b c c – a
--------------------------------------------------------------------------------------------------------------
1 31 480 481 450 7 559 840 1009 450
4 259 660 709 450 9 799 960 1249 450
6 451 780 901 450 10 931 1020 1381 450
Example 7. Find formulas for all primitive Pythagorean triples (a, b, c) for which the difference between
c and a is 2. From Theorem 31, we know that c – a = 2n2 = 2, so n2 = 1, and n = 1. Note that since n has no prime factors, there are no values for p and thus no restrictions on k. Thus, for n = 1, we have
a = (2k)2 – 1 = 4k2 – 1 (8a)
b = 2(2k) = 4k (8b)
c = (2k)2 + 1 = 4k2 + 1 (8c)
Below is a table of these triples for k = 1 through 6.
k a b c c – a
---------------------------------------------------
1 3 4 5 2
2 15 8 17 2
3 35 12 37 2
4 63 16 65 2
5 99 20 101 2
6 143 24 145 2
Now, looking at rows 2, 3, and 4 in this table, we can see a very interesting pattern. The sum of the bold-face entries in the left diagonal (from upper left to lower right) is 15 + 12 + 65 = 92, and the sum of the underlined bold-face entries in the right diagonal (from lower left to upper right) is 63 + 12 + 17, which is also 92! Thus, the sums of the entries in the diagonals of these three consecutive rows are equal!
Let’s look at another three consecutive rows; say, rows 4, 5, and 6. The sum of the entries in the left diagonal is 63 + 20 + 145 = 228, and the sum of the entries in the right diagonal is 143 + 20 + 65 = 228. Again, these two sums are equal!
As another example, look at rows 6, 7, and 8 in the table of Example 5 above. Here we see that the sum of the left diagonal entries (275 + 280 + 533 = 1088) is the same as the sum of the right diagonal entries
(435 + 280 + 373 = 1088).
Thus, it seems that for any table of primitive Pythagorean triples where c – a is a constant value, the sums of the entries in the diagonals of any three consecutive rows in these tables are equal.
To show that this pattern is always true, let’s consider any table of primitive Pythagorean triples where c – a is a constant and which is arranged in increasing order of their a-values. Let n be the number of the nth row of this table, and an, bn, and cn be the a, b, and c values of the primitive Pythagorean triples in this row. Then, the sum of the entries in the left diagonal of rows n, n+1, and n+2 is an + bn+1 + cn+2 and the sum of the entries in the right diagonal is an+2 + bn+1 + cn. We must show that these two sums are equal.
Now, c – a is a constant, so cn+2 – an+2 = cn – an. Adding an, an+2, and bn+1 to both sides of this equation, we get
cn+2 – an+2 = cn – an
cn+2 – an+2 + an + an+2 + bn+1 = cn – an + an + an+2 + bn+1
cn+2 + an + bn+1 = cn + an+2 + bn+1
an + bn+1 + cn+2 = an+2 + bn+1 + cn,
proving that these two diagonal sums are indeed equal. This gives us
Theorem 32 (Diagonal sums for constant c – a)
Let an, bn, and cn be the a, b, and c values of the primitive Pythagorean triple in the nth row of any
table of primitive Pythagorean triples (a, b, c) which is arranged in increasing order of their a-values,
where c – a is a constant. Then, the sums of the entries of the left and right diagonals of any three
consecutive rows in this table are equal; that is, an + bn+1 + cn+2 = an+2 + bn+1 + cn.
________________________________________________________________________________
Now Theorem 32 applies to any table of triples for which c – a is a constant, but two more interesting and even extraordinary patterns can be seen in the table of Example 7 shown again below for c – a = 2.
k a b c c – a
---------------------------------------------------
1 3 4 5 2
2 15 8 17 2
3 35 12 37 2
4 63 16 65 2
5 99 20 101 2
6 143 24 145 2
Notice that the sum of the a and b values of the first triple (3 + 4) plus the b value of the second triple
(3 + 4 + 8) equals the a-value of the second triple,15, and notice that the sum of the b and c values of the fourth triple (16 + 65) plus the b value of the fifth triple (16 + 65 + 20) equals the c-value of the fifth triple,101.
How nice!
In general, it seems that if ak, bk, and ck are the a, b, and c values of a triple in the kth row of this table, then
1) ak + bk + bk+1 = ak+1
and 2) bk + ck + bk+1 = ck+1 Let’s see.
From (8a), (8b), and (8c) above, we have ak = 4k2 – 1, bk = 4k, and ck = 4k2 + 1, so
ak+1 = 4(k+1)2 – 1 and ck+1 = 4(k+1)2 + 1.
Thus, ak + bk + bk+1 = (4k2 – 1) + 4k + 4(k + 1) = 4k2 + 8k + 4 – 1 = 4(k2 + 2k + 1) – 1
= 4(k + 1)2 – 1 = ak+1, proving 1),
and bk + ck + bk+1 = 4k + (4k2 + 1) + 4(k + 1) = 4k2 + 8k + 4 + 1 = 4(k2 + 2k + 1) + 1
= 4(k + 1)2 + 1 = ck+1, proving 2), and giving us
Theorem 33
Let an, bn, and cn be the a, b, and c values of the primitive Pythagorean triple (a, b, c) in the nth row
of any table of primitive Pythagorean triples arranged in increasing order of their a-values, where
c – a = 2. Then, 1) an + bn + bn+1 = an+1, and 2) bn + cn + bn+1 = cn+1.
________________________________________________________________________________
Example 8. Use Theorem 33 to construct the first 6 triples of the table in Example 7, where c – a = 2, starting with the triple (3, 4, 5). From (8b) of Example 7, we know that b = 4k for k = 1, 2, 3, . . . , so the b-values will always be consecutive multiples of 4. Let’s put them in our table to get
k a b c
--------------------------------------
1 3 4 5
2 8
3 12
4 16
5 20
6 24
Now, to find the a-value for row 2, we simply add the sum of the a and b-values from row 1 to the b-value from row 2, getting (3 + 4) + 8 = 15, and similarly, to find the c-value for row 2, we simply add the sum of the b and c-values from row 1 to the b-value from row 2, getting (4 + 5) + 8 = 17. Our table is now
k a b c
--------------------------------------
1 3 4 5
2 15 8 17
3 12
4 16
5 20
6 24
Next, to find the a-value for row 3, we simply add the sum of the a and b-values from row 2 to the b-value from row 3, getting (15 + 8) + 12 = 35, and to find the c-value for row 3, we simply add the sum of the b and c-values from row 2 to the b-value from row 3, getting (8 + 17) + 12 = 37. The table now becomes
k a b c
--------------------------------------
1 3 4 5
2 15 8 17
3 35 12 37
4 16
5 20
6 24
Continuing in this same fashion for row 4, we get (35 + 12) + 16 = 63 for the a-value, and (12 + 37) + 16 = 65 for its c-value. Following this procedure, we end up with
k a b c
--------------------------------------
1 3 4 5
2 15 8 17
3 35 12 37
4 63 16 65
5 99 20 101
6 143 24 145
c) The difference between a and b. We can write a – b as a – b = (m2 – n2) – 2mn, but this does not lead to any simple algebraic expression. The best we can do is to interchange the left and right-hand sides of this equation, getting
(m2 – n2) – 2mn = a – b,
or m2 – 2mn – n2 = a – b. Adding 2n2 to both sides of this equation, we get m2 – 2mn + n2 = a – b + 2n2,
or (m – n2 = 2n2 + a – b. Taking the positive square root of both sides of this equation, we have __________
m – n = √ 2n2 + a – b
__________
or m = n + √ 2n2 + a – b (9)
__________
Since m and n are integers, then √ 2n2 + a – b must be an integer also, requiring that 2n2 + a – b be a perfect square. But how do we find those values of n, a, and b which make this a perfect square?
To find some patterns, let’s look at Table 2 in section 3 above and calculate the various values for a – b for all the triples found there. We will omit any differences of 1 or -1 (unit differences), since we have already discussed these earlier in this section. Proceeding in order from top to bottom in Table 2 and omitting repetitions and those values greater than 300, we get from the second triple in the table, (15, 8, 17), a – b = 15 – 8 = 7, from the third triple,
(35, 12, 37), a – b = 35 – 12 = 23, and so forth. Thus, we have a – b =
7, 23, 47, 79, 79, 119, -17, 167, 223, 287, 41, -31, 73, 113, 161, 217, 281, 103, 71, 17, 151, 49, 97, 271, 89.
Since we really don’t care whether the differences are positive or negative, let’s write down just the absolute values of these differences in numerical order, giving us |a – b| =
7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, 103, 113, 119, 151, 161, 167, 217, 223, 271, 223, 271, 281, 287 (9a)
Now, are there any patterns here? Let’s look at the differences between these numbers from 7 through 113 to try and find some.
|a – b| 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, 103, 113
-------------------------------------------------------------------------------------------------------------------------------
differences 10 (6 8) (10 6 2) 22 (2 6 10 8) 6 10
14 18 26
Alas, there doesn’t seem to be any pattern in these differences! The first difference is 10, which then decreases by 4 to the second difference 6. We would have liked the second difference to be 14, an increase of 4, leading to the arithmetic sequence 10, 14, 18, 22, 26, . . . However, the sum of the second and third differences is 14, which is what we want! And the sum of the fourth, fifth, and sixth differences is 18, as desired! And the seventh difference is 22! And the sum of the eighth, ninth, tenth, eleventh differences is 26. Wow! There is a pattern! (If we try hard enough.)
Let’s rewrite the table above by eliminating the numbers 23, 41, 47, 73, 79, 89, 103, and 113, getting
|a – b| 7 17 31 49 71 97
-----------------------------------------------------------------------
differences 10 14 18 22 26
and the following table:
k 1 2 3 4 5 6
---------------------------------------------------------------
|a – b| 7 17 31 49 71 97
Using the values in this table, let’s write |a – b| as a second-degree polynomial in k; that is, for some integers d, e, and f,
|a – b| = dk2 + ek + f. (10)
From the table, if k = 1, then |a – b| = 7, and (10) becomes 7 = d + e + f. (10a)
If k = 2, then |a – b| = 17, and (10) becomes 17 = 4d + 2e + f. (10b)
If k = 3, then |a – b| = 31, and (10) becomes 31 = 9d + 3e + f. (10c)
Subtracting (10a) from (10b), we get 10 = 3d + e. (11a)
Subtracting (10b) from (10c), we get 14 = 5d + e. (11b)
Finally, subtracting (11a) from (11b), we get 4 = 2d, or d = 2.
Substituting d = 2 into (11a) gives us e = 4, and substituting d = 2 and e = 4 into (10a) gives us f = 1.
Thus, (10) becomes |a – b| = 2k2 + 4k + 1 for k = 1, 2, 3, . . . , (12)
and we have found a formula for some of the a – b differences. But what about the rest?
Writing down those remaining differences from (9a) to which (12) does not apply, we have
23, 41, 47, 71, 73, 79, 89, 103, 113, 119, 151, 161, 167, 217, 223, 271, 281, 287 (13)
Are there any patterns here? It certainly doesn’t seem so! However, to get formula (12), we used those differences which formed the arithmetic sequence 10, 14, 18, 22, . . . Now, in (13) the first difference is 23, then 41, which is a difference of 18. Perhaps we can find numbers from (13) whose differences form the arithmetic sequence 18, 22, 26, 30, . . . Let’s see. Choosing specific numbers from (13) so as to form our desired sequence, we get
|a – b| 23 41 (63) 89 119 (13a)
-----------------------------------------------------------------------
differences 18 22 26 30
where we have had to put in 63 as a value for |a – b| to get our desired sequence even though it is not in the list (9a).
Forming the following table,
k 1 2 3 4 5
---------------------------------------------------------------
|a – b| 23 41 63 89 119
let’s use the values in this table, as we did earlier in a similar manner, to write |a – b| as a second-degree polynomial in k; that is, for some integers d, e, and f,
|a – b| = dk2 + ek + f. (14)
From the table, if k = 1, then |a – b| = 23, and (14) becomes 23 = d + e + f. (14a)
If k = 2, then |a – b| = 41, and (14) becomes 41 = 4d + 2e + f. (14b)
If k = 3, then |a – b| = 63, and (14) becomes 63 = 9d + 3e + f. (14c)
Subtracting (14a) from (14b), we get 18 = 3d + e. (15a)
Subtracting (14b) from (14c), we get 22 = 5d + e. (15b)
Finally, subtracting (15a) from (15b), we get 4 = 2d, or d = 2.
Substituting d = 2 into (15a) gives us e = 12, and substituting d = 2 and e = 12 into (14a) gives us f = 9.
Thus, (14) becomes |a – b| = 2k2 + 12k + 9 for k = 1, 2, 3, . . . , (16)
Now, comparing (12) with (16), we can see another pattern emerging.
|a – b| = 2k2 + 4k + 1 = 2k2 + 4(1)k + 12
|a – b| = 2k2 + 12k + 9 = 2k2 + 4(3)k + 32
If this pattern continues, we would expect that |a – b| = 2k2 + 4(5)k + 52, and in general,
|a – b| = 2k2 + 4sk + s2 for k = 1, 2, 3, . . . and s = 1, 3, 5, . . . (17)
Now, if we use formula (17) for |a – b| in (9) above, will this help us to find primitive Pythagorean triples with constant a – b differences? Let’s see!
In (9), the relationship necessary between m and n for a given difference a – b was found to be
__________
m = n + √ 2n2 + a – b . (9)
1) If a > b, then a – b is positive, and we substitute 2k2 + 4sk + s2 for a – b in (9) above to get
___________________
m = n + √ 2n2 + 2k2 + 4sk + s2 (18)
Now the last three terms of the expression under the radical sign, 2k2 + 4sk + s2, remind us of (2k + s)2, but of course, (2k + s)2 = 4k2 + 4sk + s2, not 2k2 + 4sk + s2. We need another 2k2, and we can get it if we pick n = k.
Then, the expression under the radical sign becomes
2n2 + 2k2 + 4sk + s2 = 4k2 + 4sk + s2 = (2k + s)2, a perfect square!,
__________________
and substituting n = k into (18), we get m = n + √ 2n2 + 2k2 + 4sk + s2
_____________ ________
= k + √ 4k2 + 4sk + s2 = k + √ (2k + s)2
= k + 2k + s = 3k + s.
So, when m = 3k + s and n = k, the difference between the a and b values of (a, b, c) will be the constant value 2k2 + 4sk + s2, and a will be greater than b. (a > b).
2) If a < b, then a – b is negative, and we substitute - (2k2 + 4sk + s2) for a – b in (9) above to get
___________________
m = n + √ 2n2 – 2k2 – 4sk – s2 (19)
If, as in 1) above, we let n = k, then the expression underneath the radical sign becomes - 4sk – s2, which is
negative and thus not possible. But, what if we pick n = k + s ? Then, the expression under the radical sign
becomes
2n2 – 2k2 – 4sk – s2 = 2(k + s)2 – 2k2 – 4sk – s2
= 2k2 + 4sk + 2s2 – 2k2 – 4sk – s2 = s2, again a perfect square!
______________________
and substituting n = k + s into (19), we get m = k + s + √ 2(k + s)2 – 2k2 – 4sk – s2
___
= k + s + √ s2 = k + 2s.
So, when m = k + 2s and n = k + s, the difference between the a and b values of (a, b, c) will be the constant value - 2k2 – 4sk – s2, and a will be less than b. (a < b).
As can be seen from the formulas for m and n, if k is a multiple of s, then m and n will have a common factor of s, and (a, b, c) will not be a primitive Pythagorean triple. All of this gives us the following theorem.
Theorem 34
Let (a, b, c) be a primitive Pythagorean triple. Then the possible values for | a – b | are
given by 2k2 + 4sk + s2; that is, | a – b | = 2k2 + 4sk + s2, where s is any odd positive
integer, and k is any positive integer not a multiple of s. The generating numbers m and n
for (a, b, c) are
1) m = 3k + s and n = k, if a > b,
and 2) m = k + 2s and n = k + s, if a < b.
_________________________________________________________________________
Example 9. Find three primitive Pythagorean triples (a, b, c) for which a – b = 73. Since a – b is positive, we know that a > b, so we will use part 1) of Theorem 34 to find the values of m and n for these triples. To do this, we must first solve | a – b| = 2k2 + 4sk + s2 = 73 for s and k. Choosing odd positive integers for s, we get
for s = 1, 2k2 + 4k + 1 = 73, 2k2 + 4k – 72 = 0, and k2 + 2k – 36 = 0, which is not factorable,
for s = 3, 2k2 + 12k + 9 = 73, 2k2 + 12k – 64 = 0, and k2 + 6k – 32 = 0, which is also not factorable,
for s = 5, 2k2 + 20k + 25 = 73, 2k2 + 20k – 48 = 0, and k2 + 10k – 24 = 0, so (k + 12)(k – 2) = 0, and k = 2.
Thus, s = 5 and k = 2, and substituting these values into 1) of Theorem 34, we get m = 3(2) + 5 = 11, and n = 2,
giving us a = 112 – 22 = 117, b = 2(11)(2) = 44, and c = 112 + 22 = 125, and so (117, 44, 125) is one primitive Pythagorean triple for which a – b = 73. But, how do we find two more?
In our discussion of unit triples earlier in this section, we discovered a way to find the next unit triple after any given unit triple in a list of triples arranged in increasing order of their a values. (See Theorem 29.) Since, for unit triples, the difference between their a and b values is always constant, | a – b | = 1, perhaps Theorem 29 will apply to primitive Pythagorean triples with constant value differences between their a and b values other than 1. Let’s see.
Starting with m1 = m = 11 and n1 = n = 2, and using Theorem 29, we get
m2 = 2m1 + n1 = 2(11) + 2 = 24 and n2 = m1 = 11.
Continuing in this fashion, we get the table shown below in which the triples all have a constant difference of 73 (in absolute value) between their a and b values.
m n a b c a – b
--------------------------------------------------------------------------------
11 2 117 44 125 73
24 11 455 528 697 - 73
59 24 2905 2832 4057 73
142 59 16683 16756 23645 - 73
343 142 97485 97412 137813 73
We choose (117, 44, 125), (2905, 2832, 4057), and (97485, 97412, 137813) as our three triples since
a – b is to be positive.
Apparently, Theorem 29 does apply to primitive Pythagorean triples for which the difference between their a and b values is a constant other than 1! Let’s see if we can prove this.
Accordingly, we let A1 be any triple (a1, b1, c1) with m and n values of m1 and n1, and a1 – b1 = d, a given positive integer. Next, we let A2 be another triple (a2, b2, c2) whose m and n values are m2 = 2m1 + n1 and n2 = m1. Then,
a2 – b2 = m22 – n22 – 2m2n2
= (2m1 + n1)2 – m12 – 2(2m1 + n1)m1
= 4m12 + 4m1n1 + n12 – m12 – 4m12 – 2m1n1
= 2m1n1 + n12 – m12
= - (m12 – n12 – 2m1n1)
= - (a1 – b1) = - d.
Thus, we get
Theorem 35
If A = (a, b, c) is a primitive Pythagorean triple with m = m1, n = n1, and a – b = d,
a given positive integer, then in the triple whose m and n values are given by m = 2m1 + n1
and n = m1, a – b = - d.
__________________________________________________________________________
Example 10. Find three primitive Pythagorean triples (a, b, c) for which b – a = 41. Since b – a is positive, we know that a < b, so we will use part 2) of Theorem 34 to find the values of m and n for these triples. To do this, we must first solve | a – b| = 2k2 + 4sk + s2 = 41 for s and k. To help us find s and k easily, let’s find the values of
2k2 + 4sk + s2 for various values of s and k, putting them in Table 4 below.
Table 4
Values of |a – b| = 2k2 + 4sk + s2
for s = 1, 3, 5, . . . , 15 and k = 1, 2, 3, . . . , 7.
k
1 2 3 4 5 6 7
--------------------------------------------------------------------------
1 7 17 31 49 71 97 127
3 23 41 63 89 119 153 191
5 47 73 103 137 175 217 263
s 7 79 113 151 193 239 289 343
9 119 161 207 257 311 369 431
11 167 217 271 329 391 457 527
13 223 281 343 409 479 553 631
15 287 353 423 497 575 657 743
---------------------------------------------------------------------------
As we can see from Table 4, 2k2 + 4sk + s2 equals 41 when s = 3 and k = 2. Then, from part 2) of Theorem 34, we have m = k + 2s = 2 + 2(3) = 8, and n = k + s = 2 + 3 = 5, which gives us (39, 80, 89) as one triple for which
b – a = 41. To find more triples, we use Theorem 35 to get
m n a b c b – a
----------------------------------------------------------------------------------
8 5 39 80 89 41
21 8 377 336 505 - 41
50 21 2059 2100 2941 41
121 50 12141 12100 17141 - 41
292 121 70623 70664 99905 41
and thus (39, 80, 89), (2059, 2100, 2941), and (70263, 70664, 99905) are three triples for which b – a = 41.
(If we wish |a – b| = 41, then all five triples in the table satisfy this requirement.)
Note: In (13a) above, we put in 63 as a value for |a – b| to get a desired sequence even though it was not in the list (9a). As we can see above, 63 is indeed in Table 4 and is thus a possible value for |a – b|.
Example 11. Find three primitive Pythagorean triples (a, b, c) for which a – b = 119. Since a – b is positive, we know that a > b, so we will use part 1) of Theorem 34 to find the values of m and n for these triples. To do so, we must first solve | a – b| = 2k2 + 4sk + s2 = 119 for s and k. Looking in Table 4, we find that
2k2 + 4sk + s2 equals 119 in two different places;
1) when s = 9 and k = 1, and 2) when s = 3 and k = 5.
Then, from part 1) of Theorem 34, we have
1) for s = 9, k = 1, m = 3k + s = 3(1) + 9, m = 12, and n = k, n = 1, which gives us (143, 24, 145)
as one triple for which a – b = 119, and
2) for s = 3, k = 5, m = 3k + s = 3(5) + 3, m = 18, and n = k, n = 5, which gives us (299, 180, 349)
as another triple for which a – b = 119.
To find more triples, we use Theorem 35 to get
1) m n a b c a – b
----------------------------------------------------------------------------------
12 1 143 24 145 119
25 12 481 600 769 - 119
62 25 3219 3100 4469 119
149 62 18357 18476 26045 - 119
360 149 107399 107280 151801 119
and thus (143, 24, 145), (3219, 3100, 4469), and (107399, 107280, 151801) are three triples for
which a – b = 119.
2) m n a b c a – b
----------------------------------------------------------------------------------
18 5 299 180 349 119
41 18 1357 1476 2005 - 119
100 41 8319 8200 11681 119
241 100 48081 48200 68081 - 119
582 241 280643 280524 396805 119
and thus (299, 180, 349), (8319, 8200, 11681), and (280643, 280534, 396805) are three triples for
which a – b = 119.
There is an interesting pattern in Table 4. Here are the first three rows of Table 4, written to show the differences between each two adjacent row entries (in bold-face type).
1 2 3 4 5 6 7
---------------------------------------------------------------------------
1 7 17 31 49 71 97 127
10 14 18 22 26 30
3 23 41 63 89 119 153 191
18 22 26 30 34 38
5 47 73 103 137 175 217 263
26 30 34 38 42 46
Notice that the differences between adjacent row entries in row 1, namely, 10, 14, 18, 22, . . . , form an arithmetic sequence starting at 10 and then increasing by 4 for each succeeding difference. The same is true for rows 2 and 3 except their arithmetic sequences start at 18 and 26, each 8 units higher than the previous row’s starting value, respectively. Below is Table 4 again, showing that this pattern continues throughout the table.
Table 4 (Rows)
Values of |a – b| = 2k2 + 4sk + s2
for s = 1, 3, 5, . . . , 15 and k = 1, 2, 3, . . . , 7.
k
1 2 3 4 5 6 7
---------------------------------------------------------------------------
1 7 17 31 49 71 97 127
10 14 18 22 26 30
3 23 41 63 89 119 153 191
18 22 26 30 34 38
5 47 73 103 137 175 217 263
26 30 34 38 42 46
s 7 79 113 151 193 239 289 343
34 38 42 46 50 54
9 119 161 207 257 311 369 431
42 46 50 54 58 62
11 167 217 271 329 391 457 527
50 54 58 62 66 70
13 223 281 343 409 479 553 631
58 62 66 70 74 78
15 287 353 423 497 575 657 743
66 70 74 78 82 86
----------------------------------------------------------------------------
We can easily find more values in Table 4, if needed, by simply extending the arithmetic sequences as far as we like. For example, to extend the values in the row where s = 9, we simply add 4 to 62, the last difference in its sequence, getting 66 for the next difference in the sequence. We then add this difference to 431, the last entry in the row for s = 9, getting 431 + 66 = 497 as the next entry in this row.
9 119 161 207 257 311 369 431 (497)
42 46 50 54 58 62 (66)
Below are the first four columns of Table 4, written to show the differences between each two adjacent column entries (in bold-face type). Notice that the differences between adjacent column entries in column 1, namely, 16, 24, 32, 40, . . . , form an arithmetic sequence starting at 16 and then increasing by 8 for each succeeding difference. The same is true for columns 2, 3, and 4 except their arithmetic sequences start at 24, 32, and 40, each 8 units higher than the previous column’s starting value, respectively.
Table 4 (Columns)
Values of |a – b| = 2k2 + 4sk + s2
for s = 1, 3, 5, . . . , 15 and k = 1, 2, 3, . . . , 7.
k
1 2 3 4
------------------------------------------------------------------------------------
1 7 17 31 49
16 24 32 40
3 23 41 63 89
24 32 40 48
5 47 73 103 137
32 40 48 56
s 7 79 113 151 193
40 48 56 64
9 119 161 207 257
48 56 64 72
11 167 217 271 329
56 64 72 80
13 223 281 343 409
64 72 80 88
15 287 353 423 497
--------------------------------------------------------------------------------------
Another interesting pattern in Table 4 can be found by looking at any square block in the table, such as the 3 x 3 square shown below in red.
Table 4
Values of |a – b| = 2k2 + 4sk + s2
for s = 1, 3, 5, . . . , 15 and k = 1, 2, 3, . . . , 7.
k
1 2 3 4 5 6 7
--------------------------------------------------------------------------
1 7 17 31 49 71 97 127
3 23 41 63 89 119 153 191
d . f
5 47 73 103 137 175 217 263
s 7 79 113 151 193 239 289 343 . . .
9 119 161 207 257 311 369 431
e . g
11 167 217 271 329 391 457 527
13 223 281 343 409 479 553 631
15 287 353 423 497 575 657 743
---------------------------------------------------------------------------
If we add the two numbers at the ends of the left diagonal, call them d and g, then d + g = 73 + 257 = 330, and if we add the two numbers at the ends of the right diagonal, call them e and f, then e + f = 161 + 137 = 298. Then, (d + g) – (e + f) = 330 – 298 = 32. Not so very interesting until we realize that this is always true for any 3 x 3 square in the table. For example, in the 3 x 3 square that is at the bottom right of Table 4 above, we have d = 391, e = 575, f = 527, and g = 743, and (d + g) – (e + f) = (391 + 743) – (575 + 527) = 1134 – 1102 = 32.
This pattern is true for other sizes of square blocks, also. For example:
In the 2 x 2 square shown below in orange, we have d = 223, e = 287, f = 281, and g = 353, and (d + g) –
(e + f) = (223 + 353) – (287 + 281) = 576 – 568 = 8, which is true for all 2 x 2 squares.
In the 4 x 4 square shown below in red, we have d = 137, e = 329, f = 263, and g = 527, and (d + g) – (e + f) = (137 + 527) – (329 + 263) = 664 – 592 = 72, which is true for all 4 x 4 squares.
Table 4
Values of |a – b| = 2k2 + 4sk + s2
for s = 1, 3, 5, . . . , 15 and k = 1, 2, 3, . . . , 7.
k
1 2 3 4 5 6 7
--------------------------------------------------------------------------
1 7 17 31 49 71 97 127
3 23 41 63 89 119 153 191
5 47 73 103 137 175 217 263
s 7 79 113 151 193 239 289 343
9 119 161 207 257 311 369 431
11 167 217 271 329 391 457 527
13 223 281 343 409 479 553 631
15 287 353 423 497 575 657 743
---------------------------------------------------------------------------
Let’s put all this information into a table and try to find a general expression for (d + g) – (e + f). Accordingly, we have
Square size (d + g) – (e + f)
----------------------------------------------------------------------------------
2 x 2 8 = 8(1) = 8(12) = 8(2 – 1)2
3 x 3 32 = 8(4) = 8(22) = 8(3 – 1)2
4 x 4 72 = 8(9) = 8(32) = 8(4 – 1)2
n x n 8(n – 1)2
The data in the table suggests that if we have an n x n square, then (d + g) – (e + f) = 8(n – 1)2. (20) Let’s try to prove this. Look at the 6 x 6 square below.
k k + (n-1)
3 4 5 6 7 8 → [3 + 5 = 3 + (6-1) ]
s 7 d . . . . f
9 . .
11 . .
13 . .
15 . .
17 → [7 + 10 = 7 + 2(6-1)] e . . . . g
s + 2(n-1)
Notice that for this 6 x 6 square,
1) d and f have the same value of s (namely, 7), but the k-value of f, 8, is 5 units (6 – 1) greater than the k-value of d; namely, 3 + (6 – 1).
2) e and g have the same value of s (namely, 17), but this value of s is 10 units [ 2(5) = 2(6 – 1 ] greater than the s-value of d and f; namely, 7 + 2(6 – 1), and the k-value of g, 8, is 5 units (6 – 1) greater than the k-value of e; namely, 3 + (6 – 1).
In general, for an n x n square,
1) d and f have the same values of s, but the k-value of f is n – 1 units greater than the k-value of d; namely, k + (n – 1). Thus, if
d = 2k2 + 4sk + s2,
then f = 2(k + n – 1)2 + 4s(k + n – 1) + s2.
2) e and g have the same value of s, but it is 2(n – 1) units greater than the s-values of d and f; namely,
s + 2(n – 1) = s + 2n – 2, and the k-value of g is n – 1 units greater than the k-value of e; namely.
k + (n – 1). Thus, if
d = 2k2 + 4sk + s2,
then e = 2k2 + 4(s + 2n – 2)k + (s + 2n – 2)2,
and g = 2(k + n – 1)2 + 4(s + 2n – 2)(k + n – 1) + (s + 2n – 2)2
Now, d + g = [ 2k2 + 4sk + s2 ] + [2(k + n – 1)2 + 4(s + 2n – 2)(k + n – 1) + (s + 2n – 22 ], (21)
and e + f = [ 2k2 + 4(s + 2n – 2)k + (s + 2n – 2)2 ] + [ 2(k + n – 1)2 + 4s(k + n – 1) + s2 ]. (22)
Subtracting (22) from (21), the bold-faced italicized terms cancel, and we get
(d + g) – (e + f) = [ 4sk + 4(s + 2n – 2)(k + n – 1) ] – [ 4(s + 2n – 2)k + 4s(k + n – 1) ]
= [ 4sk + 4(sk + sn – s + 2nk + 2n2 – 2n – 2k – 2n + 2) ] – [ 4sk + 8nk – 8k + 4sk + 4sn – 4s ]
= 4sk + 4sk + 4sn – 4s + 8nk + 8n2 – 8n – 8k – 8n + 8 – 4sk – 8nk + 8k – 4sk – 4sn + 4s
= 8n2 – 16n + 8 = 8(n2 – 2n + 1) = 8(n – 1)2, which proves (20). Thus, we have
Theorem 36
For any primitive Pythagorean triple (a, b, c), |a – b| = 2k2 + 4sk + s2, where s is any odd positive
integer, and k is any positive integer not a multiple of s. Consider any table of values of |a – b|, where
the row headings are values of s = 1, 3, 5, … , and the column headings are values of k = 1, 2, 3, … .
Then, for any n x n square block in this table, (d + g) – ( e + f) = 8(n – 1)2, where d and g are the
end values of the left diagonal, and e and f are the end values of the right diagonal of the square.
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