# Solution to puzzle 54: Diophantine squares

Find all solutions to c^{2} + 1 = (a^{2} − 1)(b^{2} − 1), in integers a, b, and c.

Clearly a = b = c = 0 is one integer solution.

Also, if c = 0, a^{2} − 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, (a^{2} − 1) and (b^{2} − 1) −1 or 0, and (c^{2} + 1) 1 or 2.

Hence (a^{2} − 1) (b^{2} − 1) −1 (mod 4), and (c^{2} + 1) 1 (mod 4); that is, a, b, c are even.

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

Set a = 2a_{1}, b = 2b_{1}, c = 2c_{1}.

Then 4c_{1}^{2} + 1 = (4a_{1}^{2} − 1)(4b_{1}^{2} − 1).

Simplifying, we have c_{1}^{2} = 4a_{1}^{2} b_{1}^{2} − a_{1}^{2} − b_{1}^{2}.

Considering this equation, mod 4, if a_{1}^{2} 1 or b_{1}^{2} 1, then c_{1}^{2} −1 or −2, which is impossible.

Hence a_{1}, b_{1}, c_{1} are even.

Now set a_{1} = 2a_{2}, b_{1} = 2b_{2}, c_{1} = 2c_{2}.

Then c_{2}^{2} = 16a_{2}^{2} b_{2}^{2} − a_{2}^{2} − b_{2}^{2}.

Hence a_{2}, b_{2}, c_{2} are even.

Clearly this argument can be continued indefinitely.

Thus we have an infinite sequence of positive integers, bounded below: c > c_{1} > c_{2} > c_{3} > ... > 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 c^{2} + 1 = (a^{2} − 1)(b^{2} − 1) is a = b = c = 0.

## Generalization

Do non-zero solutions to c^{n} + 1 = (a^{n} − 1)(b^{n} − 1) exist for odd n > 2?

## Further reading

- Irrationality by Infinite Descent
- Fermat's Infinite Descent
- Infinite Descent versus Induction

Source: Original

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