InternetPerson10 wrote: Find all ordered pairs $(a, b)$ of positive integers such that $a^2 + b^2 + 25 = 15ab$ and $a^2 + ab + b^2$ is prime. Are you sure this is the exact problem ? First equation $a^2 + b^2 + 25 = 15ab$ is a Pell equation with infinitely many solutions (starting with the trivial $(1,2)$ or $(2,1)$). But it seems very difficult to identify the solutions which match the second condition. For example $(a,b)=(43261,2897)$ matches the first equation while $a^2+ab+b^2=2005233847=10861\times 184627$ and so not prime (but the two prime divisors don't seem easy to find and so it seems that the solution is not to prove that all great solutions are divisible by a same given prime).

pco wrote: Are you sure this is the exact problem ? Yes.

InternetPerson10 wrote: Find all ordered pairs $(a, b)$ of positive integers such that $a^2 + b^2 + 25 = 15ab$ and $a^2 + ab + b^2$ is prime. $17(a^2+b^2+ab)=(4a+4b+5)(4a+4b-5)+a^2+b^2+25-15ab$ $=(4a+4b+5)(4a+4b-5)$ And so $(4a+4b+5)(4a+4b-5)=17p$ for some prime $p$ and so : Either $4a+4b-5=1$, and so $a+b=\frac 32$, impossible Either $p>17$ and so $4a+4b-5=17$, and so $a+b=\frac{11}2$, impossible Either $p<17$ and so $4a+4b+5=17$, and so $a+b=3$ and $\boxed{(a,b)\in\{(1,2),(2,1)\}}$, which indeed both fit.

what is the motivation for getting $$17(a^2+b^2+ab)=(4a+4b+5)(4a+4b-5)+a^2+b^2+25-15ab?$$seems kinda random to me

Let $a+b= x$ so $x^2 -ab = p$ Therefore $x^2+25 = 17ab = 17(x^2-p)$ . We get that $17p = 16x^2 -25$ . And $17p=(4x-5)(4x+5)$ then we can check in easily. For example if $ 4x-5=17$ and $4x+5=p$ we can find x from the first equation and then we are done.