7. It is known that for some natural numbers $a, b$, the number $N=\frac{a^{2}+b^{2}}{a b-1}$ is also natural. Find all possible values of $N$.

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Answer: 5.

Solution: It is not difficult to find the solution $a=2, b=1, N=5$. Let's prove that no other $N$ is possible.

Let $\left(a_{0}, b_{0}\right)$ be a solution corresponding to the smallest value of $a^{2}+b^{2}$. Without loss of generality, we can assume that $a_{0}>b_{0}$. If $b_{0}=1$, then $N=\frac{a^{2}+1}{a-1}$, from which $a=2$ or $a=3$, and in both cases $N=5$.

Now let $b_{0}>1$. Consider the quadratic equation $x^{2}-b_{0} N x+\left(b_{0}^{2}+N\right)=0$, then $a_{0}$ is its root. By Vieta's theorem, the second root is $a_{1}=b_{0} N-a_{0}$ and it is also positive and integer. From the minimality of $a^{2}+b^{2}$, it follows that $a_{1}>a_{0}$ (they cannot be equal). Then $\left(a_{1}-1\right)\left(a_{0}-1\right) \geqslant b_{0}\left(b_{0}+1\right)$, but on the other hand, $\left(a_{1}-1\right)\left(a_{0}-1\right)=a_{1} a_{0}-\left(a_{1}+a_{0}\right)+1=b_{0}^{2}+N-b_{0} N+1$. Therefore, $b_{0}^{2}+N-b_{0} N+1 \geqslant b_{0}^{2}+b$, which is impossible for $b_{0}>1$.