solution please...can anyone show the proof...

sheryl · #1 $Old$ July 17th, 2006, 03:45 AM

Is there anyone who can show that the equation:

n^4 - n^2*m^2+m^4 = s^2, (where m, n and s are positive integers) has no integer solutions with |n| > |m| except (n^2, m^2) = (1, 0)

I will appreciate it if anyone can give at least a reference where the proof can be found.

Thanks!

ThePerfectHacker · #2 $Old$ July 18th, 2006, 11:11 AM

I am working on this.
I see two ways to approach it, one method is below using Pythagorean Triples.
---
You can state that,
$n^2-m^2=2uv$
$nm=u^2-v^2$
$s=u^2+v^2$

malaygoel · #3 $Old$ July 19th, 2006, 08:12 PM

Quote:

Originally Posted by sheryl

Is there anyone who can show that the equation:

n^4 - n^2*m^2+m^4 = s^2, (where m, n and s are positive integers) has no integer solutions with |n| > |m| except (n^2, m^2) = (1, 0)

I will appreciate it if anyone can give at least a reference where the proof can be found.

Thanks!

Just a thought.

We have to prove that there does not exist two integers n and m such that
$n^4 -n^2m^2 +m^4$ is a perfect square.
So we equalled it to $s^2$ .
We can solve the equation for $n^2$ and show that the discriminant cannot be zero.
$n^4 - m^2n^2 +(m^4-s^2)=0$
We get
$(n^2 - \frac{m^2\pm \sqrt{(m^2)^2-4(m^4-s^2)}}{2})=0$
$(n^2-\frac{m^2 \pm \sqrt{4s^2 -3m^4}}{2})=0$
Since the expression is a perfect square, $4s^2 -3m^4$ has to be zero which is not possible.
We have also the case in which $4s^2 -3m^4$ is a pefect square.

Malay