Digital ACT Math Practice Question #1238
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For any prime $p$, $x^2 - y^2 = p \implies x = (p + 1)/2 = (31 + 1)/2 = 16$.
Factor the difference of squares: $(x - y)(x + y) = 31$. Since $31$ is a prime number and $x, y$ are positive integers, the only positive integer factors of $31$ are $1$ and $31$. Because $x + y > x - y$: 1) $x + y = 31$ 2) $x - y = 1$ Addi...
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