Competition problem — Number theory

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Number theory Difficulty 4.8 AIME Prove it Bulgaria

Find all primes $p < q < r$ such that $p + q = r$ and the number $(r - p)(q - p) - 27p$ is a perfect square.

Solution

It is obvious that $p = 2$ and then $r - 2 = q$ . Hence $q(q-2) - 54 = u^2$ which can be written as $(q-1)^2 - u^2 = 55 \iff (q-1-u)(q-1+u) = 55 = 1 \cdot 55 = 5 \cdot 11$ . Hence we have the following two cases:

Case 1. $q-1-u = 1$ and $q-1+u = 55$ . Then $q = u+2$ , whence $q = 29$ and $r = 29+2 = 31$ .

Case 2. $q-1-u = 5$ and $q-1+u = 11$ . Then $q = u+6$ , whence $q = 9$ , a contradiction.

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