From the 2008-9 British Mathematical Olympiad 1:

Find all of the real values of $x$, $y$ and $z$ such that:

• $(x+1)yz = 12$
• $(y+1)zx = 4$
• $(z+1)xy = 4$.

I find this sort of cyclical simultaneous equation extremely unintuitive: there are tricks and hacks I use, but it’s all very ad hoc and I don’t have a good overriding framework for tackling them. I try tricks and see what works. So today, more than usual, I’m on the lookout for alternative solutions that make me go aha!

I had several false starts with this one, before spotting something that simplified it immensely. My solution is below the line.

### Sameness

The key to my solution is to start from the last two equations. Rearranging: $4 - xyz = zx$; and $4 - xyz = xy$.

That means $zx = xy$, so either $x=0$ (which is impossible) or $z=y$. Now we only have two equations:

• $(x+1)y^2 = 12$
• $(y+1)xy = 4$.

You may be able to spot by inspection that $x=2$ and $y=-2$ solves this; I didn’t.

Instead, I figured I’d rearrange the first equation to get $x = \frac{12-y^2}{y^2}$.

Substituting this in the second equation gives: $y(y+1)(12-y^2){y^2} = 4$, or $y^4 + y^3 - 8y^2 - 12y = 0$.

### Solving the quartic

There’s an obvious ‘solution’ of $y=0$, which is again invalid so I can divide out a factor of $y$ to leave a cubic: $y^3 + y^2 - 8y - 12 = 0$.

This is unfortunate. My usual strategy for factorising a cubic involves looking at factors of the unit term; -12 is a nasty composite with 12 reasonable numbers to check.

A sketch (or at least, the idea of a sketch) might help. The graph of this curve cuts the vertical axis at -12, so (as a positive cubic), it has at least one solution for $y > 0$.

When $y=10$, the expression has a value of 1008, so there is a solution between 0 and 10. (I was hoping for some clues from the factors of 1008, but ha! good luck with that.)

How about $y=5$? That gives 98, so there’s a solution between 0 and 5 – but more to the point, since 98 is $2\times 7^2$, it suggests $(y+2)$ as a (repeated) factor, with $(y-3)$ as the other. (This is, at this point, a hunch, but one worth checking.)

A little bit of the grid method later, it turns out that the hunch is correct: the expression does indeed factorise as $(y+2)^2(y-3)$.

There is a double solution of $(x,y,z) = (2, -2, -2)$ and a single solution of $\left(\frac{1}{3}, 3, 3\right)$.