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Problem 168

AMC 10/12, early questions

Number theory Difficulty 3.7 Prove it Junior Macedonian Mathematical Olympiad · North Macedonia

Let $p$ is a prime number and let $3p+10$ is the sum of the squares of six consecutive positive integers. Prove that $36|p-7$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

From the conditions of the problem, we have that
$3p+10=(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2+(n+3)^2=6n^2+6n+19,$
so, we have that
$3p=6n^2+6n+9,$
and
$p=2n^2+2n+3=2n(n+1)+3.$
If one of the numbers $n$ or $n+1$ is divisible with $3$ , then we have a contradiction with the condition that $p$ is a prime number. So, $n=3k+1$ . Then,
$p=2(3k+1)(3k+1+1)+3=2(3k+1)(3k+2)+3=2(9k^2+9k+2)+3=18k(k+1)+7.$
Since $k(k+1)$ is an even number, we have that $36|p-7$ .

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