Can you prove if $5^n \mid r$ and $r$ odd then $5^{n+1} \mid 2^r+3^r$? I've looked around at this, but been unable to find it. After generating prime factorizations in Maple of $2^r+3^r$ for $r>0$ it seems that this holds. It's easy enough to prove case by case. The case $n=0$ is trivial since for $r$ odd $x+y|x^r+y^r$ (I'm pretty sure it works by factoring by grouping).
For $n=1$ suppose $r>0$ is odd and divisible by 5. Then $r=20k+5$ or $r=20k+15$ for some $k\in\mathbf{N}$.
Case 1: $2^{20k+5}+3^{20k+5}=2^5(2^k)^{20}+3^5(3^k)^{20}\equiv 7(1)+18(1)$ (mod 25) $\equiv$ 0 (mod 25)
Case2: $2^{20k+15}+3^{20k+15}=2^{15}(2^k)^{20}+3^{15}(3^k)^{20}\equiv 18(1)+7(1)$ (mod 25) $\equiv$ 0 (mod 25)
This works nicely because 20 is $\phi(25)$. In fact for any power of 5, $\phi(5^{n+1})=5^n(4)$ so $r$ can be broken into two cases $5^{n}(4)k+5^{n}$ or $5^nk(4)+3(5^n)$. Since a power of 2 or 3 can't be congruent to a power of 5, raising $2^k$ and $3^k$ to $\phi(5^{n+1})$ gives 1 modulo $5^{n+1}$. But here's is where I'm stuck, showing that $2^{5^n}+3^{5^n}\equiv$ $0$ (mod $5^{n+1}$). Or maybe I'm just approaching it the wrong way? It seems that by looking at Maple data this holds for any $x$, $y$, and $x+y$ where $x+y$ is odd.
 A: I'll continue from where you stopped.
Hint: Sow by induction that $5^{n+1}|2^{5^n}+3^{5^n}$. The base $n=0$ is clear. To show the step write $2^{5^{n+1}}+3^{5^{n+1}}$ as:
$$(2^{5^n}+3^{5^n})(2^{4.5^n}-2^{3.5^n}3^{5^n}+2^{2.5^n}3^{2.5^n}-2^{5^n}3^{3.5^n}+3^{5^n})$$
By the induction hypothesis, $(2^{5^n}+3^{5^n})$ is divisibe by $5^{n+1}$. Now it suffices to show that $(2^{4.5^n}-2^{3.5^n}3^{5^n}+2^{2.5^n}3^{2.5^n}-2^{5^n}3^{3.5^n}+3^{5^n})$ is divisible by 5. To do this use the fact that $2^{5^n}\equiv -(3^{5^n})\,(mod\,5)$. This follows from the fact $2\equiv -3\,(mod\,5)$ and the fact that $5^n$ is odd.
A: Here's another proof.
Lemma: If $a\equiv b \pmod {p^k}$ then $a^p \equiv b^p \pmod {p^{k+1}}$.
Proof: Suppose $a\equiv b \pmod {p^k}$. Then $a = b + np^k$ whence we have
$$a^p = \sum_{i=0}^p\binom{p}{i}b^{p-i}n^ip^{ki}$$
Taking the above modulo $p^{k+1}$, all summands with $i>1$ are eliminated by the term $p^{ki}$. The term with $i=1$ is eliminated since $p^{k+1}\mid \binom{p}{1}p^k$ leaving only the $i=0$ term where we have $$a^p\equiv b^p\pmod{p^{k+1}}$$
which shows the result. $\square$
Now note that since $r$ is odd and $5^n\mid r$ we must have $r=(4k\pm 1)5^n$. Then
$$2^r + 3^r \equiv 2^{4\cdot 5^n \pm 5^n} + 3^{4\cdot 5^n \pm 5^n} \equiv 2^{\pm 5^n}+3^{\pm 5^n}\pmod{5^{n+1}}$$
Note that negative powers in this case simply denotes the multiplicative inverse.
Now $3\equiv -2 \pmod 5$ which implies $3^5 \equiv (-2)^5 \equiv -2^5 \pmod {5^2}$ and inductively,
$$3^{5^n} \equiv -2^{5^n}\pmod {5^{n+1}}$$
and similarly
$$3^{-5^n} \equiv -2^{-5^n}\pmod {5^{n+1}}$$
which is the desired result. 
