# Solution to puzzle 54: Diophantine squares

Find all solutions to c2 + 1 = (a2 − 1)(b2 − 1), in integers a, b, and c.

Clearly a = b = c = 0 is one integer solution.
Also, if c = 0, a2 − 1 = ±1, and so a = 0; similarly b = 0.

Without loss of generality, we now seek a solution with c > 0.
Assume such a solution exists.

Consider the equation, modulo 4.
Since the square of any integer (mod 4) is 0 or 1, we have, in mod 4, (a2 − 1) and (b2 − 1) −1 or 0, and (c2 + 1) 1 or 2.
Hence (a2 − 1) (b2 − 1) −1 (mod 4), and (c2 + 1) 1 (mod 4); that is, a, b, c are even.

A proof by a form of infinite descent on c follows.

Set a = 2a1, b = 2b1, c = 2c1.
Then 4c12 + 1 = (4a12 − 1)(4b12 − 1).
Simplifying, we have c12 = 4a12 b12 − a12 − b12.
Considering this equation, mod 4, if a12 1 or b12 1, then c12 −1 or −2, which is impossible.
Hence a1, b1, c1 are even.

Now set a1 = 2a2, b1 = 2b2, c1 = 2c2.
Then c22 = 16a22 b22 − a22 − b22.
Hence a2, b2, c2 are even.

Clearly this argument can be continued indefinitely.
Thus we have an infinite sequence of positive integers, bounded below: c > c1 > c2 > c3 > ... > 0.
But this is impossible, and therefore our original assumption that a solution exists with c > 0 is false.

Therefore the only integer solution to c2 + 1 = (a2 − 1)(b2 − 1) is a = b = c = 0.

## Generalization

Do non-zero solutions to cn + 1 = (an − 1)(bn − 1) exist for odd n > 2?