I think I have a valid proof of this question, but as my textbook doesn't provide an answer, I would appreciate some clarification.
Let $f(n)=3^n+2(17^n)$, and the question asks to show $f(n)$ is never a perfect square for positive integers n.
Using quadratic residues, we see:
$x^2(mod 5)$ is either $ 0, 1, 4 $ only.
Taking $f(n)$ and breaking it up into $3^n$ and $2(17^n)$, we can see the following modulo-periodic sequences taking each part $mod5$
The sequence of $3^n (mod5)$ is: $ 3, 4, 2, 1 ,3...$, with a period of $4$, for positive integers $n$.
The sequence of $2(17^n)$ $(mod5)$ is : $4, 3, 1, 2, 4...$, with a period of 4.
Hence, $f(n)$ $(mod 5)$ is equal to the sum of each corresponding element in each sequence, taken $(mod 5)$.
This produces the sequence: $2, 2, 3, 3, 2, 2, 3, 3, 2.....$
Hence, $f(n)$ $(mod5)$ leaves remainders of $2$ and $3$ only, and as a square $(mod5)$ leaves remainders of $0, 1, 4$ only, $f(n)$ can never be a perfect square.
Thanks.
Your method is excellent.
My immediate "first try" was to look at the result $\bmod 8$, since the number is clearly odd and all odd squares are $1 \bmod 8$.
So for positive $n$ we have $3^n\equiv \{1,3\} \bmod 8$ and $17^n\equiv 1 \bmod 8$ obviously, so $f(n)\equiv \{3,5\} \bmod 8$ which is never square.