Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Intriguing but not informative title... – lhf Dec 17 '11 at 16:34
googling (xy + 1)(yz + 1)(zx + 1) gives – 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 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.

share|cite|improve this answer
Sadly i am only able to access the first page of the article.But the concept of P(t) set was interesting – supertramp Dec 23 '11 at 10:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.