We say that a triple of integers $(x, y, z)$ is of the nice type if $x, y$, and $z$ are positive integers, with $y \geq 2$, and $x^{2}-3 y^{2}=z^{2}-3$.

a) Find a triple $(x, y, z)$ of the nice type with $x=5$ and $x=7$.

b) Show that for every $x \geq 5$ and odd, there are at least two distinct triples $\left(x, y_{1}, z_{1}\right)$ and $\left(x, y_{2}, z_{2}\right)$ of the nice type.

c) Find some triple of the nice type with $x$ even.
Solution

a) For $x=5$ and $x=7$, we have some examples of nice triples: $(x, y, z)=(5,2,4)$, $(5,3,1),(7,3,5)$, and $(7,4,2)$.

b) The particular cases from the previous item allow us to conjecture the following triples for odd $x$:

$$
(x, y, z)=(2 n+1, n, n+2) \text { and }(x, y, z)=(2 n+1, n+1, n-1)
$$

To verify that they satisfy the equation, notice that

$$
\begin{aligned}
(2 n+1)^{2}-3 n^{2} & =n^{2}+4 n+1 \\
& =(n+2)^{2}-3
\end{aligned}
$$

and

$$
\begin{aligned}
(2 n+1)^{2}-3(n+1)^{2} & =n^{2}-2 n-2 \\
& =(n-1)^{2}-3
\end{aligned}
$$

c) Considering the factorization $(x-z)(x+z)=3(y-1)(y+1)$, we can conclude that $x-z$ and $x+z$ are divisors of the right-hand side of the equation. Since $x$ is the arithmetic mean of these two divisors, this allows us to define an ordered search for possible solutions of the equation with even $x$. Choosing $y=4$, we can analyze the possible pairs of positive divisors of $3 \cdot 3 \cdot 5$:

$$
(3,3 \cdot 5),(3 \cdot 3,5) \text { and }(1,3 \cdot 3 \cdot 5)
$$

Since $x-z<x+z$, we have the cases:

$$
\left\{\begin{array} { l } 
{ x - z = 3 } \\
{ x + z = 1 5 }
\end{array} \quad \left\{\begin{array} { l } 
{ x - z = 5 } \\
{ x + z = 9 }
\end{array} \quad \left\{\begin{array}{l}
x-z=1 \\
x+z=45
\end{array} .\right.\right.\right.
$$

There are no even values for $x$ in this case. The same analysis for $y=7$ also shows that there are no solutions in this case. If $y=9$, the triple $(x, y, z)=(16,9,4)$ is a nice solution with even $x$.