Algebraic Number Theory Question

  • Thread starter putongren
  • Start date
  • #1
104
0
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
Staff Emeritus
Science Advisor
Gold Member
14,950
19
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
104
0
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
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,518
1,123
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
104
0
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
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,518
1,123
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.
 
Last edited:

Related Threads on Algebraic Number Theory Question

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
2
Views
1K
Replies
13
Views
4K
  • Last Post
Replies
1
Views
689
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
1K
Top