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Problem 1126

AMC 12 late, AIME early

Number theory Difficulty 4.8 Prove it Harvard-MIT Math Tournament · United States

$p$ and $q$ are primes such that the numbers $p+q$ and $p+7q$ are both squares. Find the value of $p$ .

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

Solution:

Writing $x^2 = p + q$ , $y^2 = p + 7q$ , we have $6q = y^2 - x^2 = (y - x)(y + x)$ . Since $6q$ is even, one of the factors $y - x$ , $y + x$ is even, and then the other is as well; thus $6q$ is divisible by $4 \Rightarrow q$ is even $\Rightarrow q = 2$ and $6q = 12$ . We may assume $x, y$ are both taken to be positive; then we must have $y - x = 2$ , $y + x = 6 \Rightarrow x = 2$ , so $p + 2 = 2^2 = 4 \Rightarrow p = 2$ also.

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