Smallest $n$ such that $125|2^n+3^n$ 
Find the smallest positive integer $n$ such that $125|2^n+3^n$  

Suppose $125|2^n+3^n$. Now, $n$ must be odd because if $n$ is even then
$$0\equiv 2^n+3^n\equiv2^n+(-2)^n\equiv 2^n+2^n\equiv2^{n+1}\pmod 5,$$
which is in contrast to hypothesis. How to go on from here?
 A: Lifting the exponent lemma is useful here.

Let $p>2$ be a prime. Let $a$ and $b$ be integers not divisible by $p$ and let $n$ be an odd positive integer. If $p\mid a+b$ then  $$v_p(a^n+b^n)=v_p(a+b)+v_p(n).$$ Here $v_p(x)$ denotes the unique integer with the property that $p^{v_p(x)} \mid x$ and $p^{v_p(x)+1} \nmid x$.

Using the lemma for $p=5$, $a=2$, and $b=3$ we find that the smallest $n$ is $25$.
A: Using Proof of $a^n+b^n$ divisible by a+b when n is odd,
$$2^n+3^n$$ will be divisible by $2+3$ iff $n$ is odd $=2m+1$(say)
$$2^{2m+1}+3^{2m+1}=2^{2m+1}+(5-2)^{2m+1}$$
$$\equiv\binom{2m+1}15\cdot2^{2m}-\binom{2m+1}25^2\cdot2^{2m-1}\pmod{5^3}$$
So, we need
$$25\mid2^{2m-1}(2m+1)(2-5m)$$ 
$$\iff25\mid(2m+1)$$ as $5\nmid2^{2m-1}(2-5m)$
A: Certainly $2^{25}+3^{25}$ is divisible by $125$. One may check now by computation that $n=25$ is the smallest such $n$.
A: As $3\equiv2^7\pmod{125}$
$$2^n+3^n\equiv2^n+(2^7)^n\equiv2^n(2^{6n}+1)\pmod{125}$$
So, we need $2^{6n}\equiv-1\pmod{125}$
Using Show that $r$ is a primitive root (mod $p^k$), $2$ is a primitive root $5^n$ for $n\ge1$
$$\implies6n\equiv\dfrac{\phi(125)}2\pmod{\phi(125)}$$
$$\iff3n\equiv25\pmod{50}$$
As $17\cdot3\equiv1\pmod{50}$
$$ n\equiv25\cdot17\pmod{50}\equiv25$$
