Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Prove it Romania

a) Prove that if $p$ , $q$ , $\sqrt{2p-q}$ and $\sqrt{2p+q}$ are integers, then $q$ is even.

b) Find out how many positive integers $p$ have the property that both $\sqrt{2p-4030}$ and $\sqrt{2p+4030}$ are integers.

Solution

a) We know that $2p - q = k^2$ , $2p + q = r^2$ , hence $r^2 - k^2 = 2q$ , with $k$ , $r$ positive integers. Then $(r - k)(r + k) = 2q$ , and the conclusion follows from the fact that $r - k$ and $r + k$ have the same parity.

b) Answer: four numbers.
With the above notations, $(r - k)(r + k) = 2 \cdot 4030 = 2^2 \cdot 5 \cdot 13 \cdot 31$ . Since $r - k$ and $r + k$ have the same parity and $r - k < r + k$ , the pair $(r - k, r + k)$ can be $(2, 4030)$ , $(10, 806)$ , $(26, 310)$ or $(62, 130)$ .
Then $r \in \{2016, 408, 168, 96\}$ and $p \in \{2030, 113, 812, 17, 1209, 753\}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.