Prove that there are no such odd numbers $x,y,z$ such that satisfy both $xy+1=a^2$, $yz+1=b^2$ and $xz+1=c^2$. And, of course, $x,y,z,a,b,c\in\mathbb Z$.

I've proven it myself but I want to see some other solutions.

Here's how I've done this myself:

Since $a,b,c$ are even, we have that $(xyz)^2=(a^2-1)(b^2-1)(c^2-1)\Rightarrow 1\equiv 3\pmod 4$.

I think there are some other ways of solving this. Thanks.

  • 3
    $\begingroup$ Yours is a nice symmetrical way. My worse way is to note that two of $x,y,z$ are congruent mod $4$, say $x$ and $y$. and therefore $xy+1\equiv 2\pmod{4}$, impossible. $\endgroup$ Jan 28 '14 at 15:48
  • 1
    $\begingroup$ I like your way a lot. $\endgroup$
    – WillO
    Jan 28 '14 at 16:00

First of all, as you said we must necessarily have $a, b \text{ and } c $ even. Moreover

$\forall x \in \mathbb{Z} ,\, x^{2} \equiv 0 \text{ or } 1 \,\,(\text{mod } 4)$

If we suppose that $x,y \text{ and } z$ are odd, we have $x, y , z \equiv 1 \text{ or } 3 \, \, (\text{mod } 4)$

However, one should note that $1 \times 1 + 1 \equiv 2 \,\, (\text{mod 4})$, $1 \times 3 + 1 \equiv 0 \,\, (\text{mod 4})$, $3 \times 3 + 1 \equiv 2 \,\, (\text{mod 4})$. As a consequence, only the combination $\left\{(1,3) \, \, (\text{mod }4) \right\}$ for the term $xy + 1$ can lead to a value equal to $0 \,(\text{mod }4)$, compatible with the parity of $a, b, \text{ and } c$.

The final argument is to notice that because of the pigeonhole principle, at least one of the three pair $(x,y) , (y,z) , (z,x)$ is such that its two terms are equal $\text{mod }4$, so that theirs products plus one can't be the square of an even number.

  • $\begingroup$ Sorry, I like OP's version better. $\endgroup$
    – vonbrand
    Aug 24 '15 at 19:08

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