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Problem 1023

AMC 12 late, AIME early

Number theory Difficulty 4.9 Prove it Croatian Mathematical Competitions · Croatia

Determine all primes $p$ for which there exist positive integers $x$ and $y$ such that
$\begin{cases} p + 1 = 2x^2 \\ p^2 + 1 = 2y^2. \end{cases}$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Subtracting the given equations we get $p(p-1) = 2(y-x)(y+x)$ .
From this we conclude
$p \mid y + x,$
because otherwise $p$ would be a divisor of $y - x$ , and $p - 1$ would be a multiple of number $y + x$ , which is impossible (we would have $p - 1 \ge y + x > y - x \ge p$ then).
Since $p > y$ (from the second equation) and $y > x$ , we have $2p > y + x$ , therefore $p = y + x$ .
It follows that $p - 1 = 2(y - x)$ . By eliminating $y$ we get $p + 1 = 4x$ . By plugging that in the first equation we easily get that the only solution is $p = 7$ ( $x = 2, y = 5$ ).

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