Distinct Possible Function Values

Let $\frac{x^2+y^2+1}{xy}=t$ and consider all pair of positive integers $(x,y)$ that satisfy the equation $\frac{x^2+y^2+1}{xy}=t$ Among all such pairs $(x,y)$ , let $(a,b)$ be pair such that $a+b$ is minimum.

Claim $a=b$ .

Proof: Assume $a>b$ . Pay attention to this equation $\frac{x^2+b^2+1}{bx}=t$ which is equivalent to $x^2-btx+b^2+1=0$ a quadratic equation in $x$ . We know that $x_1=a$ is the root of this equation. The other root can be found using Vieta's formula. That is $x_2=bt-a=\frac{b^2+1}{a}$ , so $x_2$ is positive integer. Since $a>b\geq 1$ we have $x_2=\frac{b^2+1}{a}<a$ . So we get $x_2+b<a+b$ contradicting the minimality of $a+b$ . Hence $a=b$ , we proved our claim.

Since $a=b$ we get $t=\frac{2a^2+1}{a^2}$ hence $a^2$ divide 1 it means $a=1=b$ so we get $t=3$ . We conclude that 3 is the only possible value for $\frac{x^2+y^2+1}{xy}$ .

Let $k = \frac {x^2+y^2+1}{xy}$ . For a given fixed $k$ , let's consider all possible pairs $(x,y)$ of positive integers that satisfy the equation $k = \frac {x^2+y^2+1}{xy}$ . Amongst all such pairs, let $(a, b)$ be the pair which minimizes the sum $a+b$ . (Such a solution must exist.) We will show that $a = b$ .

Suppose not. Then without loss of generality, we have $a > b$ . Consider the quadratic equation $\frac {x^2 +b^2 + 1} {xb} = k \Rightarrow x^2 - kxb + b^2 + 1 = 0$ . We already know that $x_1=a$ is a root, hence by Vieta's formulae, the other root satisfies $x_2= kb-a = \frac {b^2+1}{a}$ , which is an integer by the first equation. Also, since $a > b \geq 1$ are positive integers, thus $b^2 + 1 < a^2 \Rightarrow x_2 = \frac {b^2+1} {a} < a$ . Consider the pair $(x_2, b)$ . This satisfies $\frac {x_2 ^2 + b^2 + 1}{x_2 b} = k$ and $x_2 = \frac {b^2+1}{a} < a$ , which contradicts the minimality of $a+b$ .

Hence, we must have $a=b$ which means that $\frac {2a^2+1}{a^2}$ is an integer, thus $a^2 | 1 \Rightarrow a=1$ . Hence, $k = \frac {1+1+1}{1} = 3$ is the only possible answer. Hence, there is 1 integer value of $\frac {x^2+y^2+1}{xy}$ .