$\large \begin{cases} xy+xz = 2135 \\ xy+yz =37 \end{cases}$

Find the number of triplets of positive integers $(x,y,z)$ that satisfy the system of equations above.

1
2
3
0

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The second equation can be written as $y(x + z) = 37$ . Now as $37$ is prime and $x + z \ge 2$ we must have $y = 1$ , and thus $x + z = 37 \Longrightarrow z = 37 - x$ . The first equation now becomes

$x + xz = 2135 \Longrightarrow x + x(37 - x) = 2135 \Longrightarrow x^{2} - 38x + 2135 = 0$ ,

which has no real solutions, (and thus no positive integer solutions), since the discriminant $38^{2} - 4*2135 \lt 0$ . Thus there are $\boxed{0}$ positive integer solutions $(x,y,z)$ .