Solution to puzzle 71: Consecutive cubes and squares

Show that if the difference of the cubes of two consecutive integers is the square of an integer, then this integer is the sum of the squares of two consecutive integers.
(The smallest non-trivial example is: 8³ − 7³ = 169. This is the square of an integer, namely 13, which can be expressed as 2² + 3².)

We have (m + 1)³ − m³ = 3m² + 3m + 1 = n², for some integers m and n.
Hence 12m² + 12m + 4 = 4n², from which 3(2m + 1)² = (2n − 1)(2n + 1).

Now, 2n − 1 and 2n + 1 are relatively prime.
(By Euclid's algorithm, their greatest common divisor divides their difference, namely 2. Since both are odd, their greatest common divisor must be 1.)

Therefore we must consider two possible cases, with a and b relatively prime:

2n − 1 = 3a², 2n + 1 = b²
2n − 1 = a², 2n + 1 = 3b²

Taking the first case, we have b² = 3a² + 2.
This is impossible, as any square is congruent to 0 or 1, modulo 3.

Taking the second case, notice that a must be odd.
Setting a = 2k + 1, we have 2n − 1 = (2k + 1)² = 4k² + 4k + 1.
Hence 2n = 4k² + 4k + 2 = 2[k² + (k + 1)²].
So n = k² + (k + 1)².

Therefore, if the difference of the cubes of two consecutive integers is the square of an integer, then this integer is the sum of the squares of two consecutive integers.

Remarks

Solutions of the equation are given by (m, k² + (k + 1)²) = (x_n, y_n), where (x₀, y₀) = (0, 1), (x₁, y₁) = (7, 13), and (x_n+1, y_n+1) = (14x_n − x_n−1 + 6, 14y_n − y_n−1) for n 1.
Source: American Mathematical Monthly 57 (1950), 190.

The table below shows all non-negative solutions to (m + 1)³ − m³ = [k² + (k + 1)²]², for k 10⁹.

m	k
0	0
7	2
104	9
1455	35
20272	132
282359	494
3932760	1845
54776287	6887
762935264	25704
10626317415	95930
148005508552	358017
2061450802319	1336139
28712305723920	4986540
399910829332567	18610022
5570039304932024	69453549
77580639439715775	259204175
1080558912851088832	967363152

Solution to puzzle 71: Consecutive cubes and squares

Remarks

Further reading