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?

Solution to puzzle 54: Diophantine squares

Generalization

Further reading