<s>[INST] Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$. Let $x$ and $y$ be positive integers such that $xy \mid  x^2+y^2+1$.
Show that $$ \frac{x^2+y^2+1}{xy}= 3 \;.$$
I have been solving this for a week and I do not know how to prove the statement. I saw this in a book and I am greatly challenged. Can anyone give me a hint on how to attack the problem? thanks
 [/INST] Suppose $xy\mid x^2+y^2+1$ and let $t=\displaystyle\frac{x^2+y^2+1}{xy}$ such that $t\in\mathbb{N}$.
Construct the set,
$$S=\left\{(x,y)\in\mathbb{N}\times\mathbb{N} : \frac{x^2+y^2+1}{xy}=t\in\mathbb{N}\right\}$$
We deduce that $\displaystyle\frac{x^2+y^2+1}{xy} \ge 3$ because $\displaystyle\frac{x^2+y^2+1}{xy}<3$ implies $x^2+y^2+1 \le 2xy \le x^2+y^2$ which is clearly a contradiction. Now fix $t$ and suppose that $t>3$. Since $S\neq \varnothing$, we can choose $(a,b)\in S$ such that $a+b$ is minimal and $t>3$. WLOG assume $a\ge b>0$.
Let us consider the quadratic
$$p(w)=w^2-tbw+b^2+1=0$$
It follows that $a$ is a solution since $(a,b)\in S$ and hence satisfies the quadratic equation, that is $a$ is a root. By applying  Vieta's formulas we obtain the other root $c$. Hence $a+c=tb$ and $ac=b^2+1$. Since $c=tb-a$ we have $c\in\mathbb{Z}$. Now it remains to prove that $(a,c)\in S$.
To this end suppose $c<0$. It follows that 
$$\displaystyle 0<a^2+ac+1-3ab=a^2+ac-\frac{3c}{t}(a+c)<0$$
which is clearly a contradiction. It also immediately follows that $c\neq 0$ as this implies $b^2+1=0$. Therefore $c\in\mathbb{N}$ and $(a,c)\in S$.
Now we show that this $c$ contradicts the minimality of $a$, that is $c<a$. Suppose $c>a$ so it follows that $a+1\le c$. But from Vieta's equations we obtain $\displaystyle a+1\le c=\frac{b^2+1}{a}\le a+\frac{1}{a}$ which is impossible since this inequality holds in $\mathbb{N}$ if and only if $a=1$ and hence $a=b=1$ implying $t=3$ which contradicts our assumption that $t>3$. Therefore $c\le a$. But if $c=a$ then this implies that $\displaystyle a^2=b^2+1>\frac{9}{4}b^2$ which again is a contradiction. Hence we conclude that $c<a$ and as a result $c+b$ contradicts the minimality of $a+b$. Hence $t=3$
</s>