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. 
 A: 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 $113$.
A: Trial and error is indeed a scalable solution, since after finitely many steps the sequence $2^n+3^n \pmod{113}$ must begin repeating.  A quick bit of Maple code shows that $n=56$ is the smallest solution to the OP.

Note, by Fermat's little theorem, the 113-th step is exactly the 1st step, so the period must be a divisor of 112.
A: Note that $113$ is prime, so that $a^{112}\equiv 1$ mod $113$
This means that $a^{56}\equiv \pm 1$, and if we have the positive sign we track back to $a^{28}\equiv \pm 1$ etc down to $a^7$ where there is no square root.
Now $2^7=128\equiv 15$, $2^{14}\equiv 15^2=225\equiv -1$ and $2^{28}\equiv 2^{56}\equiv 1$
And $3^5=243\equiv 17$ so that $3^7\equiv 9\cdot 17=153 \equiv 40$. Then $3^{14}\equiv 1600\equiv 18$ so that $3^{28}\equiv 324\equiv -15$ and $3^{56}\equiv 225\equiv -1$
Whence $2^{56}+3^{56}\equiv 0$

Written up like this it is not guaranteed to work. Jack's answer explains why it is guaranteed to work in this case.
A: I think your best bet is to generate tables modulo 113 for $2^n$ and $3^n$. You can see they repeat after 112 steps. So if such $n$ exists, there will be one less than 112.
Then you should just check if two values add up to 0 mod 113.
A: 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.
