Find the sum of all positive integers $n$ for which $|2^n+5^n-65|$ is a perfect square.

Step 1: Let $f(n)=2^n+5^n-65$.
We test small positive integers directly:

• For $n=1$: $|2^1+5^1-65|=|-58|=58$ (not a square).

• For $n=2$: $|2^2+5^2-65|=|4+25-65|=|-36|=36=6^2$ (perfect square!). So $n=2$ works.

• For $n=3$: $|2^3+5^3-65|=|8+125-65|=|68|=68$ (not a square).

• For $n=4$: $|2^4+5^4-65|=|16+625-65|=|576|=24^2$ (perfect square!). So $n=4$ works.

Step 2: Now we show no solutions exist for $n\ge 5$:
For $n\ge 5$, $2^n+5^n-65>0$, so the expression is $2^n+5^n-65$.
Consider the expression modulo 3:

• If $n$ is odd: $2(-1)-2=-4\equiv 2(\mathrm{mod}3)$. Since perfect squares are only $\equiv 0$ or $1(\mathrm{mod}3)$, $n$ cannot be odd.

• Therefore, $n$ must be even. Let $n=2k$ for $k\ge 3$:

Step 3: For large $k$, the term $25^k$ dominates. Specifically:

This inequality holds strictly for $k\ge 3$ (since $4^k<2\cdot 5^k+66$).
Since the expression lies strictly between two consecutive perfect squares $(5^k{)}^2$ and $(5^k+1{)}^2$, it can never be a perfect square for $k\ge 3$ (i.e., $n\ge 6$).

Step 4: The only solutions are $n=2$ and $n=4$.
The sum of all such positive integers is: