Find all primes such that $a^2+b^2=9ab-13$. 
Let $a,b$ be primes. Find all primes such that $a^2+b^2=9ab-13$. 

Whatever I've done is from parity checking. But I can't proceed with the case when both $a,b$ are odd primes. 
I tried with some modulo-chasing but couldn't complete.
 A: Not a complete answer, just a few restrictions...
Proposition 1. For any prime $p>3 \Rightarrow 3 \mid p^2-1$ (from LFT) and $8 \mid p^2-1$ (from $(2k+1)^2 \equiv 1 \pmod{8}$). As a result $24 \mid p^2-1$.
For $a>3,b>3 \Rightarrow 24 \mid a^2-1$, $24 \mid b^2-1$
and $$24 \mid a^2 + b^2 -2=9ab-15$$
or 
$$8 \mid 3ab-5 \tag{1}$$
But, any prime $p>3$ is of the $p=4k+1$ or $p=4k+3$ form. None of $a,b$ can be of the same form at the same time:


*

*$a=4k_a+1,b=4k_b+1 \Rightarrow 3ab-5=3(4k_a+1)(4k_b+1)-5=12Q-2$ is not divisible by 4 and thus not divisible by 8.

*$a=4k_a+3,b=4k_b+3 \Rightarrow 3ab-5=3(4k_a+3)(4k_b+3)-5=12Q-22$ is not divisible by 4 and thus not divisible by 8.


As a result, either
$$a=4k_a+1,b=4k_b+3 \color{red}{\text{ or }} a=4k_a+3,b=4k_b+1 \tag{2}$$
Going further:
$$3ab-5=3(4k_a+1)(4k_b+3)-5=3(16k_ak_b+12k_a+4k_b)+4$$
reveals that both $k_a,k_b$ can't be odd or even at the same time.
We can also assume $a\leq b$ and from $a^2+b^2-9ab=-13<0$, by checking $x^2-9x+1<0$ where $x=\frac{b}{a}$, we have
$$a\leq b < 9a \tag{3}$$
