Pythagorean triples and perfect squares This problem is giving me difficulty:

Show that in any Pythagorean triple there exist at most a single perfect square

So far I've been working with the equations for primitive Pythagorean Triples (ie $x = m^2 - n^2$, $y = 2mn$, $z = m^2 + n^2$) but that hasn't really worked out and the problem doesn't necessarily require the triples to be primitive, so I'm stuck here. I will appreciate any help. 
 A: Pretty neat effort by Warren. Well, if you have two or more pythagorean triplets to be squares you get trivial diophantine with no solutions. They are: 
$r^2+s^4=t^4$ and $r^4+s^4
=t^2$. 
A: Are you familiar with the following?

The area of a Pythagorean triangle is never square.

For a proof, let $H$ be the set of $h>0$ such that there exists a 
primitive Pythagorean triangle of hypotenuse $h$ whose area is square. Suppose that 
$H\ne\emptyset$, then we can take $h_0=\min\left(H\right)$, and write 
$h_0^2=x^2+y^2$, for some $x,y>0$. One of $x$, $y$ must be even, so assume that $2\mid x$. 
The area of the Pythagorean triangle is given by $A=\frac{1}{2}xy$, and since the triple is primitive, we may write:
\begin{equation*}
  A = \frac{1}{2}xy= \frac{1}{2}\cdot 2pq\left(p^2-q^2\right)
  = pq\left(p+q\right)\left(p-q\right).
\end{equation*}
But $p$, $q$, $p+q$, $p-q$ are pairwise coprime, and their product is a 
square, so we may write:
\begin{align*}
  p &= r^2, & q &= s^2, & p+q &= t^2, & p-q &= u^2,
\end{align*}
for some $r,s,t,u>0$. Putting these together we get that:
\begin{equation*}
  t^2=u^2+2s^2\Longleftrightarrow 2s^2=\left(t+u\right)\left(t-u\right),
\end{equation*}
so $2\mid \left(t+u\right)\left(t-u\right)$. The product of two integers is 
even if and only if at least one of the integers is even. But if $2\mid\left(t+u\right)$ 
or $2\mid\left(t-u\right)$, then $t$ and $u$ must have the same parity, and it follows that 
$2\mid\left(t\pm u\right)$.
Since $\left(t^2,u^2\right)=1$, we must have 
$\left(t+u,t-u\right)=2$, and so we may write:
\begin{align*}
  t+u &= 2v_0, & t-u &= 2w_0,
\end{align*}
for some $v_0,w_0>0$ with $\left(v_0,w_0\right)=1$. Recall that:
\begin{align*}
  2s^2 &= \left(t+u\right)\left(t-u\right)= 4v_0w_0, \\
  s^2 &= 2v_0w_0.
\end{align*}
So one of $v_0$, $w_0$ must be a square, and the other must be twice a 
square. Hence $v_0=v^2$, $w_0=2w^2$, and:
\begin{align*}
  t+u &= 2v^2, & t-u &= 4w^2,
\end{align*}
or vice versa. Note that in either case, we get that $s=2vw$. Now adding and subtracting the above equations we see that:
\begin{align*}
  t &= v^2+2w^2, & \pm u &= v^2-2w^2.
\end{align*}
Consequently:
\begin{equation*}
  r^2=p=\frac{1}{2}\left(t^2+u^2\right)=v^4+4w^4,
\end{equation*}
and $\left(v^2,2w^2,r\right)$ forms a Pythagorean triangle with area 
$\left(vw\right)^2$, so $r=\sqrt{p}\in H$. But $h_0=p^2+q^2$, and clearly 
$\sqrt{p}<p^2+q^2$, so $r<h_0$, contradicting the minimality of $h_0$.

Any Pythagorean triple with more than one perfect square would satisfy the Diophantine equation $x^4+y^4=z^2$, or $x^4+y^2=z^4$, and you can find Pythagorean triangles with square areas if either of these equations have integer solutions.
