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Problem 952

AMC 12 late, AIME early

Number theory Difficulty 4.4 Prove it Berkeley Math Circle · United States

If $n$ is a natural number such that $2n+1$ and $3n+1$ are perfect squares, prove that $5n+3$ can't be a prime number.

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:
Suppose that $5n+3$ is a prime number. Let $x$ and $y$ be natural numbers such that $x^2 = 2n+1$ and $y^2 = 3n+1$ . Then
$5n+3 = 4(2n+1) - (3n+1) = 4x^2 - y^2 = (2x - y)(2x + y).$
Since $5n+3$ is a prime number and $x$ and $y$ are natural numbers, we must have $2x - y = 1$ , implying that $y = 2x - 1$ .

Since $n = y^2 - x^2$ , we have that
$x^2 = 2n + 1 = 2(y^2 - x^2) + 1 = 2((2x - 1)^2 - x^2) + 1 = 6x^2 - 8x + 3.$
Thus,
$5x^2 - 8x + 3 = 0$
and the solutions are $x = 1$ or $x = 3/5$ . Since $x^2 = 2n+1$ for a natural number $n$ , neither of these $x$ would work and we have a contradiction.

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