Solution: The heart of the solution is the Vieta jumping. We prove that $$\begin{align*} \frac{m^2}{ | m^2 + 1 - n^2 | } \end{align*}$$ is a square, and hence the result follows. For any fixed integer $k$, consider the set $$\begin{align*} S_k = \{ (x,y) \mid \frac{ x^2 }{ x^2 + 1 - y^2 } = k + 1 , \quad \text{where} \quad x, y \quad \text{are odd integers}\}. \end{align*}$$ Note that the condition for $(x,y)$ to be in set $S_k$ is the same as $$\begin{align*} \frac{ y^2 - 1 }{ x^2 + 1 - y^2 } = k. \end{align*}$$ Define $\min ^*(S_k) = (u,v)$ such that for all $(a,b) \in S_k$ we have $| a | + | b | \geq | u | + | v |$, and for securing unicity, $u, v$ are positive integers, and if $| a | + | b | = u + v$, then $| a | \geq u$. By the well ordering principle, for any $k \in \mathbb{Z}$ such that $S_k \neq 0$, the function $\min ^*$ is well-defined on $S_k$ . Note that if $\ x = y \ $, we get $k + 1= x^2$. Particularly, $\mid x^2 + 1 - y^2 \mid = 1$ is a square. Also, if $y = 1$, we get $k + 1 = 1$. Particularly, $| x^2 + 1 - y^2 | = x^2$, a square. The following section will conclude that there is no $k$ such that $y \neq 1, x$ for any double $( x, y) \in S_k$. For the sake of contradiction, assume that there exists such $k'$. We will proceed with a lemma. Lemma: $\frac{1}{2} \leq \Big| \frac{x}{y} \Big| \leq 2 \ $ for any $\ (x,y) \in S_k$. Proof: If $| y | > 2 | x |$, we get $$\begin{align*} | x^2 - (y^2 - 1) | & = y^2 - 1 - x^2 \\ & > y^2 - 1 - \big( \frac{y}{2} \big)^2 \\ & = \frac{3y^2}{4} - 1\\ & > \frac{y^2 - 1}{2}. \\ \end{align*}$$ But this implies that $$\begin{align*} y^2 - 1 - x^2 = y^2 - 1, \end{align*}$$ hence $x = 0$, contradicting the fact that $x$ is odd. If $| x | > 2 | y |$, we get $$\begin{align*} | x^2 - (y^2 - 1) | & = x^2 + 1 - y^2 \\ & > 3y^2 + 1 \\ & > y^2 - 1. \\ \end{align*}$$ So, this can't happen either. Hence, the lemma is proved. $\square$ Now, note that because of $y^2 - 1 > 0$ for $k'$ ( from the assumption ) and $$\begin{align*} \frac{y^2 - 1}{x^2 + 1 - y^2} = k', \end{align*}$$ we have $$\begin{align*} k' > 0 \iff x > y. \end{align*}$$ Let $(m, n)$ be the minimal pair with respect to the definition of $\min ^* (S_{k'})$. From the definition of $(m, n)$, we know that $m, n > 0$. We consider the cases, If $m > n$, let $d = m - n$. From $$\begin{align*} \frac{n^2 - 1}{m^2 + 1 - n^2} = k', \end{align*}$$ we get $$\begin{align*} ( n ^2 - 1 )( k' + 1 ) - k' \cdot m^2 = 0. \end{align*}$$ Using the definition of $d$, and manipulating the last expression, we get $$\begin{align} n^2 - n ( 2k'd ) - ( d^2 k' + k' +1 ) = 0. \end{align}$$ Additionally, because of the lemma, we have $2n > m > n$, implying that $0 < d < n$. Let $n, n'$ be the solutions to the quadratic equation $(1)$. By the assumption that $n \in S_{k'}$, we have $n \in 2 \mathbb{Z} + 1$. By Vieta's formulas $$\begin{align} n + n' & = 2k'd, \tag{1.1} \\ n \cdot n' & = - (d^2 k' + k' + 1). \tag{1.2} \end{align}$$ From the equation $(1.1)$, we get $n' \in 2 \mathbb{Z} + 1$. From $(1.2)$, using $n, k' > 0$, we have $n' < 0$. Hence, $$\begin{align*} | n | - | n' | = n + n' = 2k'd > 0. \end{align*}$$ Hence, $$\begin{align*} | n | > | n' |. \end{align*}$$ But this implies that we have the solution $(n' + d, n')$ with $$\begin{align*} | n' + d | + | n' | = | n' + d | - n', \\ \end{align*}$$ where $$\begin{align*} | n' + d | - n' \in \{ d, -2n' - d \}, \end{align*}$$ therefore, we have a smaller solution, $$\begin{align*} | n' + d | - n' & < 2n + d \\ & = |m| + |n|. \\ \end{align*}$$ Hence, in this case, we have a contradiction. If $m < n$, as before, we know that $m < n < 2m$ from the lemma. Hence, $n = m + d$ with $0 < d < m$. The equation $$\begin{align*} \frac{ y^2 - 1 }{ x^2 + 1 - y^2 } = k' \end{align*}$$ is equivalent to $$\begin{align*} ( y ^2 - 1 )( k' + 1 ) - k' \cdot x^2 = 0. \end{align*}$$ After reordering and substituting $y = x + d$, we get $$\begin{align*} x^2 + x( 2d )( k' + 1 ) + ( d^2 - 1 )( k' + 1 ) = 0. \end{align*}$$ Let $m, m'$ be the roots of this equation with respect to $x$, where $\min ^* ( S_{k'}) = ( m, m + d)$. From Vieta's formulas, and $k' + 1 < 0$, we get $$\begin{align*} m + m' & = -2d( k' + 1 ) > 0 \\ m \cdot m' & = (k' + 1)(d^2 - 1) < 0. \\ \end{align*}$$ The last inequality holds; since $d = n - m$ is even and $d > 0$, we have $d \geq 2$. The two inequalities, taken together, yield $m > 0 > m'$. Also, from the first we have $m' \in 2 \mathbb{Z} + 1$. Hence, $\ (m', m' + d) \in S_{k'}$. Also, $$\begin{align*} | m | - | m' | & = m + m' \\ & = -2d(k' + 1)\\ & > 0 \end{align*}$$ implying $$\begin{align*} | m | > | m' |. \end{align*}$$ However, $$\begin{align*} | m' + d | + | m' | & = | m' + d | - m' \\ & \in \{ d, -2m' - d \}. \\ \end{align*}$$ In both cases we get $$\begin{align*} d & < 2m + d, \\ \text{and} \quad -2m' - d & < -2m' \\ & < 2m \\ &< 2m + d. \\ \end{align*}$$ However, this implies that $(m', m' + d)$ is a smaller element in $S_{k'}$ , a contradiction. Hence, there is no $k$ such that $y \neq 1, x$ for any double $( x, y) \in S_k$. The only non-contradictory option was $m = n$ for $\min ^* (S_k) = ( m, n )$. From this we deduce that $k + 1 = x^2$ for all $k$ such that $S_{k}$ is nonempty. Thus $| m^2 + 1 - n^2 |$ is always a square. $\blacksquare$ Tags: Solution, Vieta jumping, symmetry, smallest element Remark: If $( x, y ) \in S_k$, then we know that $k$ is positive and $\ x^2 + 1 - y^2 > 0$. Particularly, notice that $S_k$ is empty for $k < 0$. The most important line of reasoning is that $S_k \neq \emptyset$ implies that there is a way to contradict infinite descent unless the "smallest" element of $S_k$ is of the form $( t, t )$, and that implies $k = l^2 - 1$ for some ( odd ) integer $l$. Note: Do you know why $m, n$ must be odd? What would change if at least one was even?