TITLE: Prove that no four positive integers $a, b, c $ and $d$ with $ab = 2d²$ can satisfy the equation $a² + b² = c²$.
QUESTION [6 upvotes]: Prove that :

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

Thank you...

REPLY [4 votes]: 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)