One problem we were given in our number theory course was to show that no Pythagorean triangle can have its area equal its hypotenuse.

Here is my attempt:

Let the sides of the triangle be given by: $$x=u^2-v^2, \quad y=2uv,\quad z=u^2+v^2,$$ where $u>v\geq1$, and $u,v$ are of opposite parity.

Suppose that the area of this triangle was equal to its hypotenuse. Then, $$\frac{1}{2}xy=z \implies uv(u^2-v^2)=u^2+v^2 \implies \left(\frac{u}{v}\right)^2=\frac{uv+1}{uv-1}.$$

My reasoning is that this can never be the case, since the fraction on the right hand side is of the form $(n+2)/{n}$ for $n \in \mathbb{N}$, and hence never square.

Is this reasonable?

  • $\begingroup$ I don't like dividing. First note that your description of the Pythagorean triples is not right, each side can be multiplied by an integer constant $k$. That changes the equations a little. But also note that $u$ and $v$ can be taken to be relatively prime and of opposite parity. Now looking at your equation as corrected we see that $u$ divides $u^2+v^2$ so $u$ divides $v^2$, which makes things collapse because of relative primality. $\endgroup$ – André Nicolas Nov 17 '15 at 0:01
  • $\begingroup$ @AndréNicolas: Thank you, this makes sense! $\endgroup$ – MrMazgari Nov 17 '15 at 0:37
  • $\begingroup$ Also, if we had such a triple - let's say with $a < b < c$ - then $ab = 2c$ and hence $a^2b^2 = 4c^2 = 4a^2 + 4b^2 < 8b^2$, which implies $a \leqslant 2$. $\endgroup$ – Daniel Fischer Nov 17 '15 at 0:43
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    $\begingroup$ Let's be a bit more practical about this. I think it might not be too effective to argue whether the OP's interpretation is "wrong" or some other interpretation is "right", because ultimately the arbiter is the problem poser, who presumably isn't party to this conversation. Is the problem, translated into mundane algebra, as follows? Given positive integers $a, b, c$ such that $a^2+b^2 = c^2$, prove that $ab/2 \not= c$. $\endgroup$ – Brian Tung Nov 17 '15 at 0:44
  • $\begingroup$ @BrianTung: Yes, this is an equivalent form of the question, provided that $a,b$ and $c$ satisfy the requirements to be a Pythagorean triple. $\endgroup$ – MrMazgari Nov 17 '15 at 2:00

For fun we give a proof that uses no number-theoretic machinery.

Let the sides of our Pythagorean triangle be $x,y,z$, with $z$ the hypotenuse. If the area is equal to the hypotenuse, then $\frac{1}{2}xy=z$, or equivalently $2xy=4z$. Then from $x^2+y^2=z^2$ we obtain $$(x+y)^2=x^2+y^2+2xy=z^2+4z.$$ It follows that $z^2+4z$ is a perfect square. We show this is impossible.

The first perfect square after $z^2$ is $(z+1)^2$, which cannot be equal to $z^2+4z$. The next one, $(z+2)^2$, is greater than $z^2+4z$. So if $z$ is a positive integer, then $z^2+4z$ cannot be a perfect square.

Remark: We did not use the representation theorem for Pythagorean triples. Your idea to use the theorem is good. There is a little problem. Not all Pythagorean triples are of the shape you described. For example, $(9,12,15)$ is not. To represent all triples, we need to multiply the triples you described by an arbitrary positive integer constant $k$. We can take $u$ and $v$ relatively prime and of opposite parity.

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Let $b = 3$; $c = \frac {9}{\sqrt 5}$; $a = \frac {2c} b = \frac {6} {\sqrt 5}$

Then $a^2 + b^2 = \frac {36}5 + 9 = \frac {81}{5} = c^2$

So it is possible for the area of a triangle to equal the hypotenuse.

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  • $\begingroup$ Are you missing the Pythagorean triple part? And it doesn't look like $a$ and $c$ are integers to me. $\endgroup$ – MrMazgari Nov 17 '15 at 0:34
  • $\begingroup$ Um, what Pythagorean triple part? Not once in your post did you ever mention that the right triangle had integer sides or that x,y, and z were a P. triple. You never once mentioned the word "triple". $\endgroup$ – fleablood Nov 17 '15 at 0:38
  • $\begingroup$ "One problem we were given in our number theory course was to show that no Pythagorean triangle can have its area equal its hypotenuse." THAT was your question. There is no mention of integers whatsoever. $\endgroup$ – fleablood Nov 17 '15 at 0:42
  • $\begingroup$ @fleablood: A right triangle whose sides form a Pythagorean triple is called a Pythagorean triangle. The expression is not synonymous with right triangle. See also the third paragraph of the latter link. $\endgroup$ – Lucian Nov 17 '15 at 1:43

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