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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.

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  • $\begingroup$ I'll accept your answer because its clear you put in a lot of effort, but you really shouldn't ask people to accept your answer. It makes people want to not accept it by asking $\endgroup$ – Benzne_O Apr 20 '13 at 7:24
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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.

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  • $\begingroup$ Why does it follow from $x$ and $y$ being coprime their product can be written as $2pq(p^2-q^2)$? $\endgroup$ – Meow Apr 18 '13 at 17:02
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    $\begingroup$ @Alyosha It's a result of Euclid's that if $x^2+y^2=z^2$, with $x$ and $y$ coprime, and $x$ even, then there exist coprime naturals $p>q$ with $x=2pq$, $y=p^2-q^2$, and $z=p^2+q^2$. Since the OP mentioned they were aware of this result for primitive Pythagorean triples, I neglected to justify why. $\endgroup$ – Warren Moore Apr 18 '13 at 17:07
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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 $r^4+s^4 =t^2$.

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