Ukrainian problem 2

There must be odd positive integers $x,y,z$ such that $2a-1 \; =\; bx \hspace{1cm} 2b-1 = cy \hspace{1cm} 2c-1 =az$ and hence $\begin{aligned} 8a-4 & = \; 4bx \; = \; 2x(cy+1) \; = \; xy(az + 1) + 2x \\ (8 - xyz)a & = \; 4 + xy + 2x \end{aligned}$ and hence we must have $xyz \le 7$ . Since $x,y,z$ are all odd, this means that $xyz = 1,3,5$ or $7$ .

• If $xyz = 1$ then $x=y=z=1$ , which implies that $(a,b,c) = (1,1,1)$ .
• If $xyz = 3$ then $x,y,z$ equal $1,1,3$ in some order. For none of these orders is $4 + xy + 2x$ a multiple of $5$ , which means that $a$ cannot be an integer. This case is not possible.
• If $xyz = 5$ then $x,y,z$ equal $1,1,5$ in some order. For none of these orders is $4 + xy + 2x$ a multiple of $3$ , which means that $a$ cannot be an integer. This case is not possible.
• If $xyz = 7$ then $x,y,z = 1,1,7$ in some order. With $(x,y,z) = (1,1,7)$ we obtain $(a,b,c) = (7,13,25)$ . With $(x,y,z) = (1,7,1)$ we obtain $(a,b,c) = (13,25,7)$ . With $(x,y,z) = (7,1,1)$ we obtain $(a,b,c) = (25,7,13)$ .

Thus there are $4$ triples $(a,b,c)$ , making the answer $\boxed{4}$ .