15.15. For which natural numbers $a$ and $b$ is the number $a^{2}+b^{2}+1$ divisible by $a b?$
15.15. We can assume that $a \geqslant b$. According to problem 15.14 b), the pair of numbers $(a, b)=\left(F_{2 n+1}, F_{2 n-1}\right), n \geqslant 1$, has the required property; the pair of numbers $(a, b)=(1,1)$ also does. We will show that no other pairs of natural numbers $(a, b)$, where $a \geqslant b$, have this property. We apply induction on $a$. First, note that if $a=b$, then the number $2 a^{2}+1$ is divisible by $a^{2}$, so $a=1$. Suppose that $a^{2}+b^{2}+1=k a b$, where $k$ is a natural number and $a>1$. Let $a_{1}=b$ and $b_{1}=k b-a=\frac{b^{2}+1}{a}$. Then $a_{1}^{2}+b_{1}^{2}+1=b^{2}+k^{2} b^{2}-2 k a b+a^{2}+1=k^{2} b^{2}-k a b=k b(k b-a)=k a_{1} b_{1}$, so the numbers $(a_{1}, b_{1})$ have the required property. We check that $1 \leqslant b_{1} \leqslant a_{1}<a$. Indeed, $b_{1}=\frac{b^{2}+1}{a}<\frac{b^{2}+1}{b}<b+1$. The number $b_{1}$ is an integer, so $b_{1} \leqslant b=a_{1}$. Moreover, $a_{1}=b<a$ by the condition. Therefore, by the induction hypothesis, $b=a_{1}=F_{2 n+1}$ and $k b-a=b_{1}=F_{2 n-1}$ for some $n \geqslant 0$ (we consider that $F_{-1}=1$). In this case, from the equality $a^{2}+b^{2}+1=k a b$ it follows that $k=3$, so $a=k b-b_{1}=3 F_{2 n+1}-F_{2 n-1}=2 F_{2 n+1}+F_{2 n+1}-F_{2 n-1}=F_{2 n+1}+F_{2 n+1}+$ $+F_{2 n}=F_{2 n+1}+F_{2 n+2}=F_{2 n+3}$.