<s>[INST] $abcd=6(1-a)(1-b)(1-c)(1-d)$ Positive integers $a \geqslant b \geqslant c \geqslant d$ satisfy:
$$abcd=6(1-a)(1-b)(1-c)(1-d)$$
$1.$ Prove that $d=2$ 
$2.$ Find all possible values of $a,b,c$.
What I know:
Since $x$ and $x-1$ are relatively prime for all positive integers $x$ :
$$\gcd(a,1-a)=\gcd(b,1-b)=\gcd(c,1-c)=\gcd(d,1-d)=1$$
 [/INST] We can rewrite this as:
$$\prod_{cyc} \frac{a}{a-1} =6$$
We clearly have $a \geqslant b \geqslant c \geqslant d > 1$. Also, we can see that $\frac{x}{x-1}$ is a decreasing function. If $d>2$ then, 
$$\prod_{cyc} \frac{a}{a-1} < (1.5)^4<6$$
This forces $d=2$. We now have:
$$\frac{a}{a-1} \cdot  \frac{b}{b-1} \cdot \frac{c}{c-1}=3$$
Similarly assume that $c>3$. We have $(1.33)^3<3$ which is a contradiction and we will then have $c \in {2,3}$.
If $c=3$, we will have:
$$\frac{a}{a-1} \cdot \frac{b}{b-1} = 2$$
Continuing in the similar fashion, we can force $b=3$ which yields $a=4$ and thus we have $(a, b, c, d) =(4,3,3,2)$.
If $c=2$, we will have:
$$\frac{a}{a-1} \cdot \frac{b}{b-1} =\frac{3}{2}$$
Since $\frac{a}{a-1}>1$ we have $\frac{b}{b-1}<\frac{3}{2}$ which gives $b>3$. If $b>5$, we get $(1.2)^2<1.5$. We thus need to have $3<b \leqslant 5$. Although $b=3$ doesn't give integer $a$, the other cases gives us $(a, b, c, d) =(9,4,2,2)$ and $(a, b, c, d) =(6,5,2,2)$.
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