Divisor of $3^{2n+1}+61$ I have difficulty to show the following:
If $p$ is a prime and $p^2$ divides $3^{2n+1}+61$, then $p$ must be $2$.
I appreciate any help.
 A: As Will Jagy suspected this is false.
$$17^2\mid 3^{123}+61.$$
A: First part: is easy to show that 
$$
3^{2n+1}+61\equiv 0\pmod 4
$$
why? Well $61\equiv 1 \pmod 4$ (easy to verify). And is also easy to verify that 
$$
3^{2n+1}\equiv 3\pmod 4
$$
In fact every odd number can be written as $4k+3$ or $4k+1$ with $k$ integer. Let's suppose that for some $n$ we can write ($3^n$ is odd)
$$
3^n = 4k+3
$$
now we can easily verify that
$$
3^{n+1}=3(4k+3)=4(2+3k)+1
$$
and since $3=4\cdot 0+3$ is easy to see that for every odd power we have
$$
3^{2n+1}=4k+3
$$
and therefore summing (modulus 4) with $61$ you get
$$
3^{2n+1}+61\equiv 3+1\pmod 4\equiv 0 \pmod 4
$$
Now what remains to be shown is that no other prime satisfy what the OP is asking. Working on that now ;-)
EDIT: another nice way to show that (and much shorter) is:
let's write 
$$
g(n)=3^{2n+1}+61
$$
now is easy to see that
$$
g(n+1)=3^{2n+1+2}+61=3^{2n+1}3^2+61=3^{2n+1}3^2+3^2 61-8\cdot 61 = g(n)3^2-8\cdot 61
$$
and so is also true that
$$
g(n+1)  \equiv g(n)3^2-8\cdot 61 \pmod 4 \equiv 0 \pmod 4
$$
and since is true for $n=0$ is true for all $n$.
EDIT 2: let's see if we can move on with the second part. Let's use what I just wrote. Let's suppose that exist a prime $p_2>2$ for which for some $n$
$$
g(n)\equiv 0 \pmod {p_2^2}
$$
now let's consider $n+1$ we find that (similar to what we did before)
$$
g(n+1)\equiv -8\cdot 61 \pmod {p_2^2}
$$
that is not zero. I am actually not sure that is enough... Thinking a bit about this and then I will finish to edit my answer. Of course it could be true for one $n$...
