Algebraic Number Theory Question

This is actually a Number Theory question, but requires expertise that doesn't go beyond simple algebra.

Homework Statement

Show that (1+xy)(1+zy)(1+zx) is a perfect square iff (1 + xy), (1+yz) , and (1+zx) are perfect squares.

The Attempt at a Solution

I initially tried to solve it like this

let 1 + xy = s2 , 1 + yz = t2, 1 + zx = u2. I substituted the new variables and got

(1+xy)(1+zy)(1+zx) = s2t2u2 - s2t2+t4-t2+s2

I tried to get the RHS of the equation to be in some square form, but couldn't. I thought that introducing new variables to the problem would add to the complexity of the problem, but I thought that adding new variables would give me a new way to solve the problem also.

So .. any thoughts?

The Attempt at a Solution

Hurkyl
Staff Emeritus
Gold Member
What are x,y,z,s,t,u? Are you working on the 'if' part or the 'only if' part? (or have you not gotten that far yet?) Where did that equation come from?

forgot to mention x,y,z are positive integers,

I just defined s,t,u as the number whose square is xy + 1, yz +1, zx + 1 respectively

SammyS
Staff Emeritus
Homework Helper
Gold Member
This is actually a Number Theory question, but requires expertise that doesn't go beyond simple algebra.

Homework Statement

Show that (1+xy)(1+zy)(1+zx) is a perfect square iff (1 + xy), (1+yz) , and (1+zx) are perfect squares.

The Attempt at a Solution

I initially tried to solve it like this

let 1 + xy = s2 , 1 + yz = t2, 1 + zx = u2. I substituted the new variables and got

(1+xy)(1+zy)(1+zx) = s2t2u2 - s2t2+t4-t2+s2
That last equation is obviously wrong.

(1+xy)(1+zy)(1+zx) = s2t2u2 , which is a perfect square.

That's the easier half of the proof.

Now show that if (1+xy)(1+zy)(1+zx) = w2, then each of (1+xy), (1+zy), and (1+zx) are perfect squares.

That last equation is obviously wrong.

(1+xy)(1+zy)(1+zx) = s2t2u2 , which is a perfect square.

That's the easier half of the proof.

Now show that if (1+xy)(1+zy)(1+zx) = w2, then each of (1+xy), (1+zy), and (1+zx) are perfect squares.

Yeah, I suspect it was wrong, but I still have no idea how to show that (1 + xy)(1 + zy)(1 +wz) = w2.

SammyS
Staff Emeritus