3 Equations 3 Variables

Note that $(T-x)(T-2y)(T-4z) = T^3-9T^2+26T-24 = (T-2)(T-3)(T-4).$

So $(x,2y,4z)$ is a permutation of $(2,3,4)$ . Five of these six permutations lead to solutions where at least two are integers (the exception is $(4,3,2)$ ). So the answer is $\fbox{5}$ .

Let $u = 2y$ and $v = 4z$ , then the three equations are:

$u + v + x = 9$

$uv + vx + xu = 26$

$uvx = 24$

Then $(u,v,x)$ are the roots of the equation $f(w) = w^3 - 9w^2 + 26w -24$ . And the roots of $f(w)$ are $(2,3,4)$ .

$\Rightarrow (x, 2y, 4z) = (2,3,4)$

The solutions $(x,y,z)$ with at least two of $x,y,z$ are integers are: $(2, 2, \frac {3}{4} ), (2, \frac {3}{2}, 1 ), (3, 1, 1), (3, 2, \frac {1}{2} ), (4, 1, \frac {3}{4} )$ .

The answer is $\boxed{5}$ .

Brute force can also be used. But, of course, it is quite tedious.

First, assume $x,y,z\in\mathbb Z$ and get that $\boxed{(3,1,1)}$ is the only solution (the only permutation of one of the 3-tuples $(3,1,1)$ $(3,-1,-1)$ $(-3,-1,1)$ that works (only the permutations of these can work since $xyz=3$ )).

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Assume $x,y\in\mathbb Z, z\not\in\mathbb Z$ . We then have $z=\frac{3}{xy}$ , hence $xy\not\mid 3$ .

Substitute the value of $z$ into another equation $x+2y+\frac{12}{xy}=9$ and you'll see that $xy\mid 12$ , which means there's a limited amount of possibilities for the value of $xy$ , and it's even a bit more limited due to the fact that $xy\not\mid 3$ .

This leads us to the solutions $\boxed{(2,2,\frac{3}{4})},\boxed{(4,1,\frac{3}{4})},\boxed{(3,2,\frac{3}{5})}$ .

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Assume $y,z\in\mathbb Z, x\not\in\mathbb Z$ . We have $x=\frac{3}{yz}$ , hence $yz\not\mid 3$ .

Substitute the value of $x$ and we have $\frac{3}{yz}+2y+4z=9$ , which means $yz\mid 3$ , but this contradicts $yz\not\mid 3$ , hence there are no solutions in this case.

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Assume $x,z,\in\mathbb Z, y\not\in\mathbb Z$ . We have $y=\frac{3}{xz}$ , which means $xz\not\mid 3$ .

Substitute this value of $y$ and get $x+\frac{6}{xz}+4z=9$ , which means $xz\mid 6$ , which, again, means there's a limited amount of values $xz$ can take. Going through them, we see that $\boxed{(2,\frac{3}{2},1)}$ is the only solution in this case.

Therefore, there are $5$ solutions (all the ones I've boxed). $\square$