# Question about the proof that the area of right triangle cannot be square.

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 .

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?

• @DirkLiebhold Sorry, I want to prove this by proof by contradiction. – glimpser Jul 10 '17 at 7:18
• 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... – Dirk Jul 10 '17 at 7:20
• @DirkLiebhold This comes from rational points on unit circle. so one can make the numerator and the denominator relatively prime. – glimpser Jul 10 '17 at 7:23
• 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. – Dirk Jul 10 '17 at 7:25
• @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 – Henry Jul 10 '17 at 7:27

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 $gcd(u+v,u-v)=2$.