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If 1 were congruent number... 
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#1
Mar2011, 01:14 AM

P: 3,243

I want to show that if 1 were a congruent number then there would be an integer solution to the equation x^4y^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^42^4= (X^2Y^2)^2 I would like to show that X^2Y^2 is my odd number, but don't see how. Thanks for any hints. 


#2
Mar2011, 03:10 AM

P: 688

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


#3
Mar2011, 06:04 AM

P: 3,243

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



#4
Mar2011, 09:03 AM

P: 688

If 1 were congruent number...
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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