If 1 were congruent number

  • #1
MathematicalPhysicist
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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.
 

Answers and Replies

  • #2
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A hint is: read about the properties of primitive pythagorean triples. By the way, if the definition of "congruent number" is this one,
then X,Y,Z are rational, and you want integers; you are probably forgetting to mention a step in your proof.
 
  • #3
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I am not even sure if this is a proof, this is why I am asking for help here.
 
  • #4
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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!
 

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