Is 1 a Congruent Number with Odd Integer Solutions to x^4 - y^4 = u^2?

[MathematicalPhysicist]

I want to show that if 1 were a congruent number then there would be an integer solution to the equation x^4-y^4=u^2 where u is odd.

Not sure, but from the definition we have 1=XY/2 and X^2+Y^2= Z^2, so by adding and substracting 4 (2XY) I get (X+-Y)^2 = Z^2+-4
multiply them both to get: Z^4-2^4= (X^2-Y^2)^2
I would like to show that X^2-Y^2 is my odd number, but don't see how.

Thanks for any hints.

 

[dodo]

A hint is: read about the properties of primitive pythagorean triples. By the way, if the definition of "congruent number" is this one,

http://en.wikipedia.org/wiki/Congruent_number​

then X,Y,Z are rational, and you want integers; you are probably forgetting to mention a step in your proof.

 

[MathematicalPhysicist]

I am not even sure if this is a proof, this is why I am asking for help here.

 

[dodo]

It looks good so far; in fact, you're almost there.

Let me put an example: if you have an equation on fractions, say 1/3 + 1/6 = 1/2, you can multiply it by some number, and get an equation on integers. (i'm afraid to say much more, short of solving it myself.) Go ahead!

 

[blue_raver22]

I would approach this problem by first defining congruent numbers and understanding their properties. A congruent number is a positive rational number that can be represented as the area of a right triangle with rational side lengths. In other words, if n is a congruent number, then there exists positive rational numbers a, b, and c such that n = (a*b)/2 and a^2 + b^2 = c^2.

Now, let us assume that 1 is a congruent number. This means that there exist positive rational numbers a, b, and c such that 1 = (a*b)/2 and a^2 + b^2 = c^2. We can then rewrite this as a^2 - c^2 = b^2, which is a Pythagorean triple.

Next, we can use the fact that every Pythagorean triple can be generated by the formula a = m^2 - n^2, b = 2mn, and c = m^2 + n^2, where m and n are positive integers with m > n.

Substituting this into our equation, we get (m^2 - n^2)^2 - (m^2 + n^2)^2 = (2mn)^2. Simplifying, we get (m^4 - n^4) - (m^4 + 2m^2n^2 + n^4) = 4m^2n^2. Rearranging, we get (m^4 - 2m^2n^2 + n^4) - (m^4 - n^4) = 4m^2n^2. This can be rewritten as (m^2 - n^2)^2 - (m^2 - n^2)^2 = 4m^2n^2, which simplifies to 0 = 4m^2n^2.

Since m and n are positive integers, this means that at least one of them must be even. Without loss of generality, let's say m is even. This means that m^2 is divisible by 4, and therefore 4m^2n^2 is divisible by 4. This implies that (m^2 - n^2)^2 = 0, which means that m^2 - n^2 = 0.