Find all triples of positive integers (x, y, z)

Question

I've no idea how to attempt such a question. I tried eliminating z from the equations to get a quadratic equation in terms of x, so I could use the fact that the discriminant would be a perfect square. Please help

Answer 1

$$(x,y,z) = (-3,6,z)\quad {\rm or}\quad (12,13,z)$$

for any integer value of $z$.

So there are an infinite number of solutions.

Answer 2

You have the $2$ equations

$$x^2 + y - z = 100 \tag{1}\label{eq1A}$$

$$x + y^2 - z = 124 \tag{2}\label{eq2A}$$

Next, \eqref{eq2A} minus \eqref{eq1A} gives

$$\begin{equation}\begin{aligned} x - x^2 + y^2 - y & = 24 \\ -4x^2 + 4x + 4y^2 - 4y & = 96 \\ -(4x^2 - 4x + 1) + (4y^2 - 4y + 1) & = 96 \\ (2y - 1)^2 - (2x - 1)^2 & = 96 \\ (2y - 2x)(2y + 2x - 2) & = 96 \\ (y - x)(y + x - 1) & = 24 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Also, as Neat Math's comment states, it would have been simpler & easier to use $x - y + y^2 - x^2 = -(y - x) + (y - x)(y + x) = (y - x)(y + x - 1)$.

With $x$ and $y$ being positive integers means $y + x - 1 \gt 0$, so $y - x \gt 0$. Next, note the parity of $y - x$ and $y + x - 1$ are different, i.e., one is even and the other odd. The only odd & even set of divisors of $24$ are $(1, 24)$ and $(3, 8)$. Since $y + x - 1 \gt y - x$, this means we have first the possibility of

$$y - x = 1 \tag{4}\label{eq4A}$$

$$y + x - 1 = 24 \implies y + x = 25 \tag{5}\label{eq5A}$$

Adding \eqref{eq4A} and \eqref{eq5A} gives $2y = 26 \implies y = 13$, so $x = 12$, with \eqref{eq1A} and \eqref{eq2A} giving $z = 57$. The second possibility gives

$$y - x = 3 \tag{6}\label{eq6A}$$

$$y + x - 1 = 8 \implies y + x = 9 \tag{7}\label{eq7A}$$

Adding \eqref{eq4A} and \eqref{eq5A} gives $2y = 12 \implies y = 6$, so $x = 3$, with \eqref{eq1A} and \eqref{eq2A} giving $z = -85$. However, since this is not a positive integer, it's not allowed.

Thus, the only triple of positive integers solution is $(x, y, z) = (12, 13, 57)$.