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Let $a^2+b^2=c^2$ be a Pythagoras equation for some right triangle with hypotenuse = c

we can assume that $a=2pq, b=p^2-q^2, c=p^2+q^2$ (where p, q are relatively prime and have opposite parity e.g (odd,even) or (even, odd).) and assume that its area is a square.

Then its area is $ab/2=2pq*(p^2-q^2)/2=pq(p^2-q^2)=pq(p+q)(p-q)$

So $pq(p+q)(p-q)$ should be square.

By our assumption, if $p,q$ are relatively prime and one of which is even, $p+q, p-q$ are also relatively prime.

with this condition, each of $ p, q, p+q, p-q$ should be square.

Then let $p=x^2, q=y^2, p+q=u^2, p-q=v^2$ (WLOG assuming $ p>q$).

Since $p, q$ have opposite parity, $u^2, v^2$ are odds. Therefore $u, v$ are odds and relatively prime.

(1) Then it becomes $u^2=v^2+2b^2 \rightarrow 2y^2=u^2-v^2=(u+v)(u-v)$

(2) $ u,v$ are odds $\rightarrow $ $u+v, u-v$ are both even.

Then define another number $r=(u+v)/2, s=(u-v)/2$

So $r^2+s^2=(\frac{u+v}{2})^2+(\frac{u-v}{2})^2 =\frac{u^2+v^2}{2}=\frac{2p}{2}=p=x^2$

It implies that there's a right triangle with the base = $r$, the height = $s$, the hypotenuse=$x.$

At the same time, $rv=\frac{(u+v)(u-v)}{2}=\frac{y^2}{2}=\frac{q}{2}.$

$q$ is a square and $rv$ is an integer. Therefore $q/2$ is even. It follows that $\frac{rv}{2}=\frac{q}{4}$ is an integer. Besides it is a square.

So we can always make another smaller right triangle .

This is a contradiction.

Therefore the area of right triangle cannot be a square.

Here are points which I couldn't understand

At (1), someone said $gcd(u+v,u-v)=2$. But in my opinion gcd is not 2, but $y$. Could you explain why this is 2?

And at (2), among $u+v, u-v$, he claimed that one of which is divisible by 4. But I cannot find cue. Could explain it to me?

And above my argument, are there any errors? If they are, could you point out?

Thank you for your nice answer in advance.

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  • $\begingroup$ @DirkLiebhold Sorry, I want to prove this by proof by contradiction. $\endgroup$ – glimpser Jul 10 '17 at 7:18
  • $\begingroup$ Sorry, already noticed this and deleted my comment. Thus, here is a new remark: You are assuming $p$ and $q$ to be relatively prime. Why? There are many triangles out there which do not fulfill this condition, and it looks like you want a proof for all of them... $\endgroup$ – Dirk Jul 10 '17 at 7:20
  • $\begingroup$ @DirkLiebhold This comes from rational points on unit circle. so one can make the numerator and the denominator relatively prime. $\endgroup$ – glimpser Jul 10 '17 at 7:23
  • $\begingroup$ Yes, for the primitive triples. However, you are talking about the area of triangles (with integer side lengths), not about the triples. If you are trying to show a property of primitive triples and want to use the area of triangles to do so, this is not really clear from your question. $\endgroup$ – Dirk Jul 10 '17 at 7:25
  • $\begingroup$ @DirkLiebhold: multiplying $p$ and $q$ by some integer $n$ will multiply the sides by $n^2$ and the area by $n^4$, so if the area cannot be a square in the primitive case, then it cannot be a square more generally $\endgroup$ – Henry Jul 10 '17 at 7:27
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Let $gcd\left(\frac{u+v}{2},\frac{u-v}{2}\right)=d$.

Hence, $\frac{u+v}{2}+\frac{u-v}{2}=u$ divided by $d$ and $\frac{u+v}{2}-\frac{u-v}{2}=v$ divided by $d$.

Thus, $d=1$ and $gcd(u+v,u-v)=2$.

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  • $\begingroup$ Thank you for your answer! $\endgroup$ – glimpser Jul 10 '17 at 7:30
  • $\begingroup$ @Grimza You are welcome! $\endgroup$ – Michael Rozenberg Jul 10 '17 at 7:34

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