The sum of the primes, p, that satisfy the condition that $8^p+15^p$ is a perfect square. Well, the question is so:
Suppose P is the set of all primes, p, that satisfy the condition that $8^p+15^p$ is a perfect square.
Find the sum of the elements of P.
Now, over here, I found out a few bits of information. However, I've no idea how to use them, and whether they can even be used.
For example, it should be obvious that 2 belongs to P. As $64+225 = 289 = 17^2$
Also,
$15\equiv -8 \pmod{23}$
So, for all primes greater than 2, (which are obviously odd)
$8^p+ 15^p\equiv 8^p+ (-8)^p\equiv 8^p-8^p \equiv 0 \pmod{23}$
Suppose $8^p+15^p = x^2$, then, $x^2 \equiv 0 \pmod{23}$,
And, as 23 is a prime, this means that, $x \equiv 0 \pmod{23}$ and $x^2 \equiv 0 \pmod{23^2}$.
I also found out that,
$8^p +15^p \equiv 8^{p-(p-1)}+15^{p-(p-1)} \pmod{p}$, by Euler's Theorem.
Therefore,
$x^2\equiv 8^p+15^p \equiv 8+15 \equiv 23 \pmod{p}$.
Hmm. I'd really appreciate it if, instead of giving me a solution flat out, someone could point me in the right direction and perhaps offer a slight push.
Thanks a lot, people.
 A: Consider what perfect squares are modulo 5 (if you don't know that, work out the possibilities) and then consider the equation $8^p + 15^p$ mod 5. $15^p$ should be easy, and what possibilities are there for $8^p$?
A: Let $v_p(n)$ denote the maximum prime power in $n$. That is if, $v_p(n)=\alpha\implies p^{\alpha}\mid n$ but $p^{\alpha+1}\nmid n$.
As, $23\mid 8+15$
By Lifting the Exponent Lemma:
$v_{23}(8^p+15^p)=v_{23}(8+15)+v_{23}(p)=1+v_{23}(p)$.
For, odd prime $p$, $23\mid 8^p+15^p$ but $23^2\mid 8^p+15^p\iff p=23$.
So, answer $=2+23=25$.
A: Note that for any $n$, we have $n^2\equiv 0\pmod3$ or $n^2\equiv 1\pmod3$. (To see it, check for each of the cases $n=3k$, $n=3k+1$, $n=3k-1$.)
Since $3|15$, then $3|15^m$ for all $m>0$, so $3|15^p$ for all (positive) primes $p$, and so $15^p\equiv 0\pmod3$ for any prime $p\in P$. On the other hand, $3$ fails to divide any power of $8$, so $8^p\not\equiv 0\pmod3$ for any prime $p\in P$.
With those observations (and the fact that $2\in P$) in mind, we'll be able to determine precisely what the set $P$ is.
Assume, then that $p\in P$, so that $8^p+15^p=n^2$ for some $n$. We cannot have $n^2=0\pmod3$, for then $8^p\equiv 8^p+15^p\equiv 0\pmod3$, which we already determined to be false. Thus, we must have $n^2\equiv 1\pmod3,$ and so $$8^p\equiv 8^p+15^p\equiv 1\pmod3.\tag{1}$$ Writing $8=9-1$, binomial expansion gives us $$8^p=\sum_{k=0}^p\binom{p}{k}(-1)^k9^{p-k}=(-1)^p+3\sum_{k=0}^{p-1}\binom{p}{k}(-1)^k3^{2p-2k-1},$$ so $8^p\equiv(-1)^p\pmod3,$ and combining this with $(1)$ gives us $$(-1)^p\equiv 1\pmod3.\tag{2}$$ Since $-1\not\equiv 1\pmod3$, then $p$ cannot be odd.
Hence, $P=\{2\}$.
