Consecutive Squares Equations

David W. Hansen

From the Pythagorean Theorem, we know that 3² + 4² = 5². Notice that the three squares are in

consecutive order with two squares on the left and one square on the right. We call this a consecutive

squares equation. Are there others? Let’s see.

Since a² + b² = c², then for the squares to be consecutive, we must have

b = a + 1 and c = a + 2, (1)

giving us a² + (a + 1)² = (a + 2)²,

or a² + (a² + 2a + 1) = a² + 4a + 4.

Subtracting a² from both sides of this equation and rewriting it with all terms to the left,

we get a² – 2a – 3 = 0.

Factoring this, we get (a – 3)(a + 1) = 0,

so a = 3 or a = - 1.

Substituting into (1), we get a = 3, b = 4, c = 5 or a = - 1, b = 0, c = 1.

If we require that all integers be positive, then we see that there is only one consecutive

squares equation; namely, 3² + 4² = 5².

But what if we take more squares, say

a² + (a + 1)² + (a + 2)² = (a + 3)² + (a + 4)²? (2)

Then a² + (a² + 2a + 1) + (a² + 4a + 4) = (a² + 6a + 9) + (a² + 8a + 16),

or a² – 8a – 20 = 0.

Factoring and solving for a, we get (a – 10)(a + 2) = 0, so a = 10 or a = - 2.

Substituting a = 10 into (2) gives us: 10² + 11²+ 12² = 13² + 14² and we have found

another consecutive squares equation! So far we have

3² + 4² = 5²

10² + 11²+ 12² = 13² + 14²

This suggests a pattern for consecutive squares equations. In the first equation above, there are two

squares to the left and one square to the right of the equals sign. In the second, there are three squares to

the left and two squares to the right. Let’s try four squares to the left and three to the right; namely,

a² + (a + 1)² + (a + 2)² + (a + 3)² = (a + 4)² + (a + 5)² + (a + 6)², (3)

or a² + (a²+2a+1) + (a²+4a+4) + (a²+6a+9) = (a²+8a+16) + (a²+10a+25) + (a²+12a+ 36)

4a² + 12a + 14 = 3a² + 30a + 77,

or a² - 18a - 63 = 0.

Factoring, we have (a + 3)(a – 21) = 0, so a = 21 or a = - 3.

Substituting a = 21 into (3), we get 21² + 22²+ 23² + 24² = 25² + 26² + 27²,

another consecutive squares equation. Very fascinating! We now have

3² + 4² = 5²

10² + 11²+ 12² = 13² + 14²

21²+ 22²+ 23² + 24² = 25² + 26² + 27²

To distinguish between these equations, let’s call the number of squares on the right-hand side of a

consecutive squares equation, the order of the equation. Thus, the last equation above is of order 3

because it has three squares on its right-hand side.

Now, look at the first terms on the left-hand side of each of the equations above; namely, 3, 10, and 21.

If we can find a pattern in these starting numbers, we can find the starting number for the consecutive

squares equation of order 4.

Let’s set up a table which lists the factors of each of these starting numbers along with the order of the

equation in which they appear.

Table 1

Starting Expanded Number of squares on

Order Number Factors Factors Left Right

1 3 1(3) 1[2(1) + 1] 2 1

2 10 2(5) 2[2(2) + 1] 3 2

3 21 3(7) 3[2(3) + 1] 4 3

4 -- -- -- 5 4

Now if the pattern for the factors of each starting number in Table 1 continues, we would expect the factors

for the equation of order 4 to be 4(9), making the starting number 36, and that there would be five squares

on the left and four squares on the right. We would then get

36² + 37²+ 38² + 39² + 40² = 41² + 42² + 43² + 44²

or 1296 + 1369 + 1444 + 1521 + 1600 = 1681 + 1764 + 1849 + 1936

and 7230 = 7230.

So, this is true! How very amazing! We now have

3² + 4² = 5²

10² + 11²+ 12² = 13² + 14²

21² + 22²+ 23² + 24² = 25² + 26² + 27²

36² + 37²+ 38² + 39² + 40² = 41² + 42²+ 43² + 44²

Let’s rewrite Table 1 to include this new result and see if we can’t find a general formula for the starting

numbers of all consecutive squares equations.

Table 1 (amended)

Starting Expanded Number of squares on

Order Number Factors Factors Left Right

1 3 1(3) 1[2(1) + 1] 2 1

2 10 2(5) 2[2(2) + 1] 3 2

3 21 3(7) 3[2(3) + 1] 4 3

4 36 4(9) 4[2(4) + 1] 5 4

k k(2k + 1) k + 1 k

Now, by looking at the patterns of the first four starting numbers in Table 1 above, we surmise that the

factors of the starting number F of the consecutive squares equation of order k will be k and 2k + 1. Thus, it

appears that F = k(2k + 1), and there will be k + 1 squares on the left-hand and k squares on the right-hand

side of the equation. Its two sides would then be

Left-hand Side (k + 1 squares):

F² + (F + 1)² + (F + 2)² + (F + 3)² + . . . + (F + k)²

Term 1 Term 2 Term 3 Term 4 Term k+1

Right-hand Side (k squares):

(F + k + 1) ² + (F + k + 2)² + (F + k + 3)² + . . . + (F + k + k)²

Term 1 Term 2 Term 3 Term k

Now if we can show that these two sides are equal, then we will have found a general expression for the

consecutive squares equation of order k. Let’s see if we can.

The left-hand side becomes

F² + (F² + 2F(1) + 1²) + (F² + 2F(2) + 2²) + . . . + (F² + 2F(k) + k²)

= F² + [kF² + 2F(1 + 2 + . . . + k) + (1² + 2² + . . . + k²)]. (4)

The right-hand side becomes (letting G = F + k)

(G + 1)² + (G + 2² + (G + 3)² + . . . + (G + k)²

= (G² + 2G(1) + 1²) + (G² + 2G(2) + 2²) + . . . + (G² + 2G(k) + k²)

= kG² + 2G(1 + 2 + . . . + k) + (1² + 2² + . . . + k²) (5)

Replacing G by F + k in (5), we have

k(F + k)² + 2(F + k)(1 + 2 + . . . + k) + (1² + 2² + . . . + k²)

= k(F² + 2Fk + k²) + 2F(1 + 2 + . . . + k) + 2k(1 + 2 + . . . + k) + (1² + 2² + . . . + k²)

= kF² + 2Fk² + k³ + 2F(1 + 2 + . . . + k) + 2k(1 + 2 + . . . + k) + (1² + 2² + . . . + k²)

= 2Fk² + k³ + 2k(1 + 2 + . . . + k) + [kF² + 2F(1 + 2 + . . . + k) + (1² + 2² + . . . + k²)] (6)

Comparing (6) with (4), we see that the three terms in the brackets in both expressions are identical,

whereas the first three terms of (6) differ from the remaining first term of (4). Perhaps we can show that

these first three terms of (6) equal the first term of (4).

Since 1 + 2 + . . . + k = k(k+1)/2, the first three terms of (6) become

2Fk² + k³ + 2k(1 + 2 + . . . + k)

= 2Fk² + k³ + 2k [ k(k + 1)/2 ]

= 2Fk² + k³ + (k³ + k²)

= 2Fk² + 2k³ + k²

= 2Fk² + k²(2k + 1)

= 2Fk² + k[k(2k + 1)]

Since F = k(2k + 1), we have = 2Fk² + kF

= kF(2k + 1)

= F[k(2k+1)]

= F², which is the first term of (4)!

Replacing the first three terms in (6) by F², we have

F² + [kF² + 2F(1 + 2 + . . . + k) + (1² + 2² + . . . + k²)], (7)

which is identical to (4) above. Thus, the left-hand side (4) and the right-hand side (7) of the consecutive

squares equation of order k are identical, and we have proved the following:

Consecutive Squares Equations

In a consecutive squares equation of order k, there are k + 1 consecutive squares

on the left-hand side and k consecutive squares on the right-hand side.

The first term is F², the last term is (F + 2k)², where F = k(2k+1), and the equation is

F² + (F+1)² + (F+2)² + . . . + (F+k)² = (F+k+1) ² + (F+k+2)²+ . . . + (F+2k)²,

or F² + . . . + F_L² = F_R² + . . . + (F+2k)²,

where F_L (the base of the last term on the left-hand side) = F + k, and

F_R (the base of the first term on the right-hand side) = F + k + 1 = F_L + 1.

Example 1. Find the consecutive squares equation of order 6. Since the order is 6, then k = 6,

F = (2k+1) = 6(13) = 78, and F + 2k = 78 + 2(6) = 90. Thus, the first term is 78², and the last term

is 90². F_L = F + k = 78 + 6 = 84, and F_R = F_L + 1 = 85. There are k+1 = 7 squares on the left

and k = 6 squares on the right side of the equation, giving us

78² + 79² + 80² + 81² + 82² + 83² + 84² = 85² + 86² + 87² + 88² + 89² + 90².

Example 2. Find the order of the consecutive squares equation whose third term is 57². Since its third

term is 57², its first term F² must be 55². Thus F = k(2k+1) = 55, or 2k² + k – 55 = 0,

(2k + 11)(k – 5) = 0, k = - 11/2 or k = 5. The order is thus 5, and the equation is

55² + 56² + 57² + 58² + 59² + 60² = 61² + 62² + 63² + 64² + 65².

Example 3. Find a consecutive squares equation with 15 terms. Since there are k+1 terms on the left-

hand side and k terms on the right-hand side of the equation, there will be a total of (k+1) + k = 2k + 1

terms. So, 2k + 1= 15, and k = 7. Thus, the order is 7. So, k = 7, F = k(2k+1) = 7(15) = 105,

F_L = F + k = 105 + 7 = 112, F_R = F_L + 1 = 113, F + 2k = 105 + 2(7) = 119, giving us

105²+ 106²+ . . . + 112² = 113² + 114² + . . . + 119².

Example 4. Find a consecutive squares equation such that the base of its last term is 20 more than the

base of its first term. Since the base of the last term is F + 2k and the base of the first term is F, we have

F + 2k = F + 20, 2k = 20, k = 10. So, F = k(2k+1) = 10(21) = 210, F_L = F + k = 210 + 10 = 220,

F_R = 221, and F + 2k = 210 + 2(20) = 230, giving us

210²+ 211² + . . . + 220² = 221² + 222² + . . . + 230².

Example 5. Find the 13^th term of a consecutive squares equation of order 8. Since the order is 8, then

k = 8, and the base of the first term F = k(2k+1) = 8(17) = 136. The base of the second term will be one

greater than the fi rst term; namely, 137. The base of the third term will be two greater than the first term;

namely, 138, and so on. Thus, the base of the 13^th term will be 12 greater than the first term; namely,

136 + 12 = 148, and the 13^th term will be 148².

References

[1] Ball, W. W. Rouse, Mathematical Recreations & Essays, American ed., New York: The Macmillan

Company, 1947, p. 58

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