There is no Pythagorean triple in which the hypotenuse and one leg are the legs of another Pythagorean triple. According to Wikipedia,

There are no Pythagorean triples in which the hypotenuse and one leg are the legs of another Pythagorean triple.

I cannot find the proof in the citation provided. I am really struggling to proving this, can somebody please direct me to a proof (or give some advice on how to prove it, or prove it themselves). Thanks!
 A: Oh, well. I cannot find this claim in Wikipedia... one case is easy, the other, not. Any primitive Pythagorean triple is $2xy, x^2 - y^2, x^2 + y^2$ with one of $x,y$ odd and the other even, and $\gcd(x,y) =1.$ We cannot have a Pythagorean triple with legs $x^2 - y^2, x^2 + y^2,$ because we would be solving $$ 2 x^4 + 2 y^4 = z^2  $$ with one of $x,y$ odd. Since that gives us $2 x^4 + 2 y^4 \equiv 2 \pmod 4$ we find immediately that this cannot be an integer square.
the other case is harder, and I do not have it yet. With legs $2xy, x^2 + y^2,$ we are solving $$ \color{magenta}{ x^4 + 6 x^2 y^2 + y^4 = z^2 \; \; \; ?}   $$ My little computer run suggests that this is impossible. However....anyway, Fermat's method of descent, which I am not getting to work, is said to be equivalent to something or other using elliptic curves. So i am putting this here and adding that as a tag to the main question. 
I think I have the important step, and that there is an elementary book somewhere on elliptic curves with the necessary final step. Mostly from Mordell's book, page 139, we divide the equation in magenta by $y^4,$ replace $x/y$ by $X$ and $z^2/y^4$ by $Z^4,$ after which we are asking about the rational points on
 $$ \color{magenta}{ X^4 + 6 X^2  + 1 = Z^2 \; \; \; ?}   $$
Mordell points out that the points, and especially the rational points, given by the equations
$$  S = \frac{X^2 + Z + 1}{2}  $$
and then
$$ T = 2 X (S+1)  $$ take us to the elliptic curve
$$ T^2 = 4 S^3 - 4 S,  $$ which has no constant term. It is my fervent prayer that this elliptic curve has only the trivial rational points at which $T=0,$ so that the only rational points on the original have $X=0,$ and finally, back to lower case, either $x=0$ or $y=0.$ 
I WILL FIND OUT 
See rational points on particular elliptic curve
A: Suppose that (i) $a^2+b^2=c^2$ and (ii) $b^2+c^2=d^2$. Then, if we divide $(i)$ by $b^2$ we obtain $u^2+1=v^2$ and dividing (ii) by $b^2$ obtain $1+v^2=w^2$, where $u=a/b$, $v=c/b$ and $w=d/b$. In particular $u^2+1=v^2$, $u^2+2=w^2$, and $u^2$ are all squares. Hence, $(u^2,uvw)$ is a point in the elliptic curve 
$$E:y^2=x(x+1)(x+2).$$
After a change of variables, $x= X-1$, we see that $E$ is isomorphic to 
$$Y^2=X^3-X=X(X-1)(X+1).$$
This curve only has $4$ rational points, namely $\{\infty,(0,0),(1,0),(-1,0)\}$, so the only points on $E$ are $\{\infty,(0,0),(-1,0),(-2,0)\}$, and therefore the only solutions of (i)+(ii) are trivial.
