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Show that $17$ is the only prime of the form $p^q +q^ p$ , where $p$ and $q$ are prime

My attempt so far is first assume $p$ and $q$ are prime. Now $17=2^3+3^2.$

Now fix $p=2$ and let $q>3$ then $q=3x+1$ or $q=3x+2$, $x \in \mathbb{Z}$

then $2^{3x+1}+(3x+1)^2=9x^2+6x+2(2^{3x})+1$

Now I'm not really sure what to do from here my goal was to show its not prime from both cases of $q$ but I think I"m missing the algebra necessary to show that.

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If $p,q$ are both odd (or both even) then $p^q+q^p$ is even and $>2$, hence nt prime. So if $n=p^q+q^p$ is prime we can assume wlog that $q=2$ and $n=p^2+2^p$ with $p$ odd. Then $2^p\equiv 2\pmod 3$.But if $p$ is not a multiple of $3$, then $p^2\equiv 1\pmod 3$ and so $2^p+p^2$ is a multiple of $3$. As $2^p+p^2$ cannot be equal to $3$, the only valid remaining case is $p=3$, $n=2^3+3^2=17$.

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    I haven't gone over mods yet for this book so is there a way to do it without that use as well. Is the general idea that $2^p$ when divided by $3$ has a remainder of two and $p^2$ has a remainder of $1$ then I see right away when added together you get a multiple of $3$. I don't see yet why those two will have that remainder. – 2017-02-04
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    @HighSchool15: $2^p \equiv (-1)^p\pmod{3}$ and $(3k\pm 1)^2\equiv 1\pmod{3}$. – 2017-02-04
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    Or $2^p-2=2^{2k+1}-2=2\cdot(4^k-1^k)=2\cdot (4-1)\cdot(4^{k+1}+4^{k-2}+\ldots+1)$ and $(3k\pm1)^2=9k^2\pm 6k+1=3\cdot(3k^3\pm 2k)+1$ – 2017-02-04