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?