# $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square if and only if each factor is

Show that if $x, y, z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square if and only if $xy + 1, yz + 1, zx+1$ are perfect squares.

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Intriguing but not informative title... – lhf Dec 17 '11 at 16:34
googling (xy + 1)(yz + 1)(zx + 1) gives jstor.org/stable/2691347 – sdcvvc Dec 17 '11 at 17:01
@sdcvvc, maybe you can summarize the contents of that paper here so that the question doesn't remain unanswered. – J. M. Dec 18 '11 at 3:53

Here's the brief summary of the article posted by sdcvvc. It is essentially a proof by descent, showing that if you had a triple $(x,y,z)$ such that the product $(xy+1)(yz+1)(xz+1)$ was a square with one of the three factors not a square, then you could find a smaller such triple (ordered, say, via the sum $x+y+z$.)
The descent is rather direct: If $(x,y,z)$ is a triple with $x\leq y\leq z$, then so is $(x,y,z')$, where $$z'=x+y+z+2xyz-2\sqrt{(xy+1)(xz+1)(yz+1)}$$ (Recall that the term under the square root was assumed square.) Their remains some checking to do; namely, that this is indeed such a triple, and that that $0<z'<z$, but this is all rather straight-forward.
Lest this seem entirely ad hoc, let me just note that, as I learned from Kedlaya's article, that sets of this type (with the property that pairwise products are of a fixed distance from a square...in our case we are learning about sets $\{x,y,z\}$ with each pairwise product one less than a square) have been heavily studied by Fermat, Diophantus, and a slew of more modern mathematicians, featuring some applications of Baker's theory of linear forms in logarithms. I'd recommend taking a look at the original article -- it's brief, informative, and well-written.