# 2021 JMPSC Invitationals Problems/Problem 13

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Let $p$ be a prime and $n$ be an odd integer (not necessarily positive) such that $$\dfrac{p^{n+p+2021}}{(p+n)^2}$$ is an integer. Find the sum of all distinct possible values of $p \cdot n$.

## Solution

Assume temporarily that $p \neq 2$. Then, $p$ and $n$ are both odd which implies that the numerator is odd and the denominator is even. Since an even number cannot divide an odd number, we have a contradiction. Since our claim is incorrect, $p=2$ and we now wish to make $$\frac{2^{n+2023}}{(n+2)^2}$$ an integer. We see that the denominator is always odd, while the numerator is always even. Thus, the only possible values for $n$ are when the denominator is $1,-1$, which implies $n=-1,-3$. These correspond with $p=2$, so $pn=-2,-6$ for an answer of $-8$. ~samrocksnature

## Solution 2

Suppose $p$ is odd. We have that our expression can never be an integer, so $p=2$. Now, $\frac{2^{n+2023}}{(2+n)^2}$ is an integer, implying $(2+n)$ is a perfect power of $2$. Now, we have $n$ is even if $(2+n) \ge 1$, and $n=\{-1,-3 \}$ are the only values that work when testing for odd $n$. The answer is $2(-1-3)=\boxed{-8}$

~Geometry285