1. Find all ordered pairs $(a, b)$ of positive integers such that $a^{2}+b^{2}+25=15 a b$ and $a^{2}+a b+b^{2}$ is prime.
Solution. The only solutions are $(1,2)$ and $(2,1)$.
First, note that $a \neq b$; otherwise we have $13 a^{2}=25$ which is impossible. We can assume without loss of generality that $1 \leq a2$. Note that $17\left(a^{2}+a b+b^{2}\right)=16 a^{2}+32 a b+16 b^{2}-25=(4 a+4 b-$ 5) $(4 a+4 b+5)$. With the condition that $a \geq 3, b \geq 4$, we have $4 a+4 b+5>4 a+4 b-5 \geq$ $23>17$, that is, $17\left(a^{2}+a b+b^{2}\right)$ is the product of two positive integers greater than 17. Thus, $a^{2}+a b+b^{2}$ is the product of two positive integers greater than 1 , and so is composite.
This tells us that indeed $(1,2)$ and $(2,1)$ are the only solutions.