Positive Integers to make a square How many integers $n$  make the expression $7^n + 7^3 + 2\cdot7^2$ a perfect square?
Factoring $7^2$ we have that $7^n + 7^3 + 2\cdot7^2 = 7^2\cdot(7^{n-2} + 9)$.
How do we prove that the 2nd factor only has $1$ solution when $n = 3$?

Given $7^n+7^3+2\cdot7^2$
Note that $7^3+2\cdot7^2=441=21^2$
So, our equation becomes,
$$7^n+441=p^2\mbox{ for some integer p} $$$$7^n=p^2-441$$$$7^n=(p+21)(p-21)$$
Since $7$ is a prime number, $(p+21)$ and $(p-21)$ should be degrees of $7$
So, the only possible value of $n$ is $3$
