Find a possible n such that $(2^n +3^n)/{113}$ is integer Find an $n$ such that $(2^n +3^n)/{113}$ is an integer.

So essentially, $2^n + 3^n$ has to become a multiple of $113$ for some $n$.

I have tried to solve it algebraically, but it is impossible because of the different bases of the exponents. Currently I have tried trial and error, but it is not a scalable solution.

I am hoping there is an elegant solution to this problem.

Thanks in advance for all your help.
Since $p=113$ is a prime number and $113\equiv 1\pmod{8}$ while $113\equiv 2\pmod{3}$, we have that $2$ is a quadratic residue in $\mathbb{F}_{113}$ while $3$ is not a quadratic residue. In terms of Legendre symbols, this gives:
$$\left(\frac{2}{p}\right)+\left(\frac{3}{p}\right) = 0, $$
hence with the choice $n=\frac{p-1}{2}=56$ we have that $2^n+3^n$ is for sure a multiple of $113$.
Though this is not a simpler way but we can solve the question using this...

It is better to draw the graphs of $$f(x)=113x$$ and $$g(x)=2^x+3^x$$ 

and find where they meet

If they don't meet there is no solution.

But, if they meet then there must be a solution and if there is an integer solution.

Find the values of each function for different integer values closer to the point of intersection and observe whether they are equal at some integer point or not.

Since both are increasing functions only one point of intersection can be seen if present.