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Prove that :

No four positive integers $a, b, c$ and $d$ with $ab = 2d²$ can satisfy the equation $a² + b² = c²$.

Thank you...

yoda
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  • It may probably be related to this parameterization of Pythagorean triples. – Lucian Feb 13 '14 at 10:44
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    If there were such integers, then there would be an integer right triangle with area a perfect square (since the area is $ab/2$). That's a famous problem; another term for it is, you are trying to prove that 1 is not a "congruent number". Fermat found a proof by "infinite descent". So, there are some search terms for you. See also http://math.stackexchange.com/questions/71946/did-leonardo-of-pisa-prove-n-4-case-of-flt and also http://math.stackexchange.com/questions/607612/problems-on-congruent-number-generating-and-others – Gerry Myerson Feb 13 '14 at 11:08
  • @GerryMyerson Thank you. – yoda Feb 13 '14 at 11:29

1 Answers1

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If such $a,b,c,d$ existed, the fact that $(a,b,c)$ is a Pythagorean triple yields integers $u,v$ such that $a=u^2-v^2,b=2uv$ (say). Putting $g={\sf gcd}(u,v)$ and $x=\frac{u}{g},y=\frac{v}{g}$, we have $a=g^2(x^2-y^2),b=2g^2xy$, so that $2g^4xy(x^2-y^2)=ab=2d^2$. Then $d'=\frac{d}{g^2}$ is an integer and

$$ xy(x-y)(x+y)=(d')^2 \tag{1} $$

Since $x$ and $y$ are coprime, they are also coprime to $x-y$ and $x+y$ (those last two may not be coprime though), so they must both be squares. Write $x=p^2,y=q^2,d''=\frac{d'}{pq}$. Then $p,q,d''$ are all integers and

$$ (p^2-q^2)(p^2+q^2)=(d'')^2 \tag{2} $$

This is a well-known impossible equation, with a classical proof by infinite descent (see Wikipedia)

Ewan Delanoy
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