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

\left|2^n+5^n-65\right|$has unit digit same as$2^n$i.e.,$2,4,8,6, \ldots \ldots$for$n=1,2,3,4, \ldots$Also for perfect square it shall be 4 or 6 Hence$n$can be even only for$n=2 \quad\left|2^2+5^2-65\right|=36 n=4 \quad\left|2^4+5^4-65\right|=576$Also$\left(K^2+1\right)^2-K^2=2 K+1$or$K^2+(2 K+1)=(K+1)^2$Let$\left|\frac{5^n}{K^2}+\frac{2^n-65}{2 K+1}\right|$to be$(K+1)^2$i.e., for$n=6 K=\left(5^3\right) \Rightarrow 2 K+1=251$but$2^6-65=-1$for$n=8 K=5^4 \Rightarrow 2 K+1=626$but$2^8-65 \Rightarrow 191$for$n=10 K=5^5 \Rightarrow 2 K+1=3126$but$2^{10}-65 \Rightarrow 959$The difference increases as$2$\left(5^x\right)+1$ rises faster than $2^x-65$ Hence only $n=2$ and 4 are possible