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 all perfect squares. Find infinitely many triples (a,b,c) of positive integers such that a, b, c are in arithmetic progression and such that $ab + 1, bc +1, ca+1$ are perfect squares.

First of all I tried to prove that $xy +1, yz +1, zx+1$ were relatively prime. I didn't succeed.

$$\gcd(xy+1, yz+1) = \gcd(y(x-z)+1,yz+1)$$

For the second problem, I have discovered the solution $(2,4,12)$ which isn't in arithmetic progression.

I have tried googling it but didn't find an easy and satisfactory solution.

Any help would be appreciated.

  • $\begingroup$ @labbhattacharjee My question also includes another question to find them. It's not a duplicate. The other question doesn't include this in case you haven't noticed. $\endgroup$ Mar 19, 2016 at 10:15
  • $\begingroup$ @Macavity The first question has not been properly answered and what are Pell equations? $\endgroup$ Mar 19, 2016 at 11:52

1 Answer 1


First question is a well known problem studied by Euler and some variants of it still out of our reach . For every positive integers $(x,y,z)$ let $P(x,y,z)$ denote the predicate: $(xy+1)(yz+1)(zx+1)$ is a square and not all of $(xy+1)$, $(yz+1)$,$(zx+1)$ are squares

Proof by infinite descent:

we well prove by the infinite descent that if there exists a solution $(x,y,z)$ such that $P(x,y,z)$ is true than there exists another solution $(p,q,r)$ such that $P(p,q,r)$ is true and $p+q+r$ less than $x+y+z$.

Let $(x,y,z)$ be tuple of positive integers such that $P(x,y,z)$ is true and $x\leq y \leq z$, consider the two positive integers $s_{+}$ and $s_{-}$ defined by : $$s_{\mp}=x+y+z+2xyz\mp\sqrt{(xy+1)(yz+1)(zx+1)} \tag 1$$ This integers verify : $$x^2+y^2+z^2+s_{\mp}^2-2(xy+yz+zx+xs_{\mp}+ys_{\mp}+zs_{\mp})-4xyzs_{\mp}-4=0 \tag2$$ And we can also prove the following important identities (they are basically the same): $$ \begin{align} (x+y-z-s_{\mp})^2&&=&& 4(xy+1)(zs_{\mp}+1) \tag 3 \\ (x+z-y-s_{\mp})^2&&=&& 4(xz+1)(ys_{\mp}+1) \tag 4 \\ (x+s_{\mp}-z-y)^2&&=&& 4(yz+1)(xs_{\mp}+1) \tag 5 \end{align}$$ we can use this identities to proof that $P(x,y,s_{\mp})$ holds, by multiplying $(4)$ and $(5)$ you will get that $(xy+1)(ys_{\mp}+1)(xs_{\mp}+1) $ is a square and not all of $(xy+1)$,$(xs_{\mp}+1)$ , $(ys_{\mp}+1)$ are square (as we have $(xz+1)$ is a square iff $xs_{\mp}+1)$ is a square and $(yz+1)$ is a square iff $(ys_{\mp}+1)$ is a square using $(4)$ and $(5)$).

Now the important par is to prove that either $x+y+s_{+}<x+y+z$ or $x+y+s_{-}<x+y+z $ this is equivalent tp proving that $s_{-}s_{+}<z^2$ which is true because : $$s_{-}s_{+}=x^2+y^2+z^2-2(xy+yz)-4=z^2-x(2z-x)-y(2z-y)-4<z^2 $$ (remember that $x\leq y \leq z$)

Reference: When Is (xy + 1)(yz + 1)(zx + 1) a Square? Kiran S. Kedlaya Mathematics Magazine Vol. 71, No. 1 (Feb., 1998), pp. 61-63

Second Question: The following Pell equation have infinitely many solutions : $$x^2-3y^2=1 $$

For any solution $(n,m)$ of this Pell equation, let $(a,b,c)=(2n-m,2n,2n+m)$ then $ab+1,bc+1,ca+1$ are all squares.


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