Problem 1692

32. First, prove a lemma: If $\alpha, \beta, \gamma \in \mathbf{N}^{*}$, such that $\alpha \beta+1, \beta \gamma+1$, and $\gamma \alpha+1$ are all perfect squares, then there exists $\delta \in \mathbf{N}^{*}$, such that the product of any two of $\alpha, \beta, \gamma, \delta$ plus 1 is a perfect square.

First, consider the conditions that $\delta$ should satisfy. Let
$$\left\{\begin{array} { l } { \alpha \delta + 1 = x ^ { 2 } , } \\ { \beta \delta + 1 = y ^ { 2 } , } \\ { \gamma \delta + 1 = z ^ { 2 } , } \end{array} \quad \left\{\begin{array}{l} \alpha \beta+1=u^{2}, \\ \beta \gamma+1=v^{2}, \\ \gamma \alpha+1=w^{2}, \end{array}\right.\right.$$

Then $\alpha \beta \gamma \delta=\left(x^{2}-1\right)\left(v^{2}-1\right)=\left(z^{2}-1\right)\left(u^{2}-1\right)$, i.e.,
$$\begin{aligned} (x v-z u)(x v+z u) & =x^{2}-z^{2}+v^{2}-u^{2}=\alpha \delta-\gamma \delta+\beta \gamma-\alpha \beta \\ & =(\alpha-\gamma)(\delta-\beta) . \end{aligned}$$

Clearly, when $(x v-z u, x v+z u)=(\alpha-\gamma, \delta-\beta)$, this equation holds. At this time, $2 x v = \alpha + \delta - \beta - \gamma$, i.e.,
$$2 \sqrt{(\alpha \delta+1)(\beta \gamma+1)} = (\alpha + \delta - \beta - \gamma)$$

Squaring and rearranging, we get
$$\delta^{2} - 2(\alpha + \beta + \gamma + 2 \alpha \beta \gamma) \delta + \alpha^{2} + \beta^{2} + \gamma^{2} - 2(\alpha \beta + \beta \gamma + \gamma \alpha) - 4 = 0$$

Solving this quadratic equation in $\delta$, we can take
$$\delta = \alpha + \beta + \gamma + 2 \alpha \beta \gamma - 2 \sqrt{(\alpha \beta + 1)(\beta \gamma + 1)(\gamma \alpha + 1)},$$

We claim that this $\delta$ satisfies the requirements of the lemma.
In fact, $\delta$ being an integer is obvious, and (13) is equivalent to $4(\alpha \delta + 1)(\beta \gamma + 1) = (\alpha + \delta - \beta - \gamma)^{2} \geqslant 0$, so the roots of (13) are all positive integers. Furthermore, from this equation, we know that $\alpha \delta + 1$ is a perfect square. Symmetrically, (13) can also be transformed into the following equations:
$$\begin{array}{l} 4(\alpha \delta + 1)(\beta \gamma + 1) = (\alpha + \delta - \beta - \gamma)^{2}, \\ 4(\beta \delta + 1)(\gamma \alpha + 1) = (\beta + \delta - \gamma - \alpha)^{2}, \\ 4(\gamma \delta + 1)(\alpha \beta + 1) = (\gamma + \delta - \alpha - \beta)^{2}, \end{array}$$

Therefore, $\alpha \delta + 1, \beta \delta + 1, \gamma \delta + 1$ are all perfect squares.
Returning to the original problem. If $(\alpha \beta + 1)(\beta \gamma + 1)(\gamma \alpha + 1)$ is a perfect square, but $\alpha \beta + 1$, $\beta \gamma + 1$, and $\gamma \alpha + 1$ are not all perfect squares. We set $(\alpha, \beta, \gamma)$ to be the set of positive integers that minimizes $\alpha + \beta + \gamma$, and assume without loss of generality that $\alpha \leqslant \beta \leqslant \gamma$.

Now, take $\delta$ as described in the lemma, and we know that $\delta$ is a positive integer. At this time, using (14), (15), and (16), we know that $\alpha \delta + 1, \beta \delta + 1, \alpha \beta + 1$ are not all perfect squares. From (14) $\times$ (15), we get
$$\begin{aligned} & 16(\alpha \beta + 1)^{2}(\alpha \delta + 1)(\beta \delta + 1)(\beta \gamma + 1)(\gamma \alpha + 1) \\ = & (\alpha \beta + 1)^{2}(\alpha + \delta - \beta - \gamma)^{2}(\beta + \delta - \gamma - \alpha)^{2} \end{aligned}$$

Combining with the fact that $(\alpha \beta + 1)(\beta \gamma + 1)(\gamma \alpha + 1)$ is a perfect square, we know that $(\alpha \beta + 1)(\alpha \delta + 1)(\beta \delta + 1)$ is also a perfect square. However,
$$\begin{aligned} \delta \cdot \delta^{\prime} & = \alpha^{2} + \beta^{2} + \gamma^{2} - 2(\alpha \beta + \beta \gamma + \gamma \alpha) - 4 \\ & < \gamma^{2} - \alpha(2 \gamma - \alpha) - \beta(2 \gamma - \beta) < \gamma^{2} \end{aligned}$$

Here, $\delta < \delta^{\prime} = \alpha + \beta + \gamma + 2 \alpha \beta \gamma + 2 \sqrt{(\alpha \beta + 1)(\beta \gamma + 1)(\gamma \alpha + 1)}$. Thus, $\delta < \gamma$, which contradicts the minimality of $\alpha + \beta + \gamma$.

The proposition is proved.