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.


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 $(y+2)^2(y-3)$.

Final answer

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

Both of these work in the original equations (and match up with the answer in the book, which is a relief.) However, I still don’t have a good feel for what’s gone on here beyond a bit of number crunching. Is there some elegance I’m missing? Is there a neat way?

Please do let me know in the comments!