Let $p$ be a prime number such that $3p+10$ is a sum of squares of six consecutive positive integers. Prove that $p-7$ is divisible by $36$.
Problem
Source: Macedonia JBMO TST 2017, Problem 1
Tags: number theory, prime numbers, Divisibility
Devastator
26.06.2018 05:13
Amir Hossein wrote: $3p+10$ is a sum of six consecutive positive integers. No such p exists since 3p+10 must be a multiple of 3 because the sum of any 6 consecutive integers is a multiple of 3 Probably just a typo, I guess Wait now that I see the corrected version, I think I remmeber seeing this a few months ago... Edit: I did see this a few months ago, I remember now Here's my sol
Let those 6 positive integers be $k,k+1,...,k+5$. Then simplifying, we get $3p+10=k^2+(k+1)^2+...+(k+5)^2=6k^2+30k+55$, so solving, we get $p=2k^2+10k+15$. Therefore, $p-7=2k^2+10k+8=2(k+1)(k+4)$
Clearly, p=3 doesn't work since the sum of the squares of 6 consecutive positive integers is clearly greater than 33, so p must not be divisible by 3. So checking mod 3, we see that $k$ must be $2mod3$ for $2k^2+10k+15$ to not be divisible by 3. This implies that $k+1$ and $k+4$ are multiples of 3, so $p-7$ is divisible by 9.
Also, since $k+1$ and $k+4$ have different parities, one of them must be divisible by 2, so $p-7$ is divisible by 4.
Therefore, $p-7$ is divisible by 36
$QED$
Amir Hossein
26.06.2018 05:16
Of course. It must be sum of squares of six consecutive integers. Fixed. Thank you!