Find all positive integers $n$ such that $7^n + 147$ is a perfect square.
Problem
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Tags: number theory, Perfect Square
math31415926535
15.09.2021 16:26
First notice that $n=0, 1$ does not work, so now lets assume $n \ge 2,$ we can factor the expression as $$7^2(7^{n-2}+3),$$since $7^2$ is a perfect square, we must also have $7^{n-2}+3$ to be a perfect square. If $n\ge 2,$ we will get that is $3\pmod{7},$ but $3$ is not a perfect square $\pmod{7},$ and $n=2$ gives $196,$ therefore, the only solution is $n=2.$
DottedCaculator
15.09.2021 17:00
If $n=0$ or $n=1$, then $7^n+147=148,154$, which are not perfect squares. Now, if $n\geq2$, then $7^n+147=49(7^{n-2}+3)$. Therefore, $7^{n-2}+3$ must be a perfect square. Either $n=2$ or $7^{n-2}+3\equiv3\pmod7$. Since $3$ is not a quadratic residue mod $3$, we must have $n=2$, so since $7^2+147=196=14^2$, the only solution is $\boxed{n=2}$.