Algebraic Number Theory Question

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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.


Homework Equations





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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Hurkyl
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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?
 
  • #3
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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
 
  • #4
SammyS
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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.


Homework Equations





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.
 
  • #5
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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.
 
  • #6
SammyS
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Yeah, I suspect it was wrong, but I still have no idea how to show that (1 + xy)(1 + zy)(1 +xz) = w2.
Did you try multiplying the left side out? -- Then try refactoring it.
 
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