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Problem

I am working on a problem to find all triangles whose sides are $a,b,c\in \mathbb{Q}$,and its area $S=1$

My attempt

By using Heron’s fomula, $$ S=\sqrt{\frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}}\\=1$$ and $$x+y> z \mbox{, for} \left\{x,y,z\right\}=\left\{a,b,c\right\}.$$

Then,to deal with the complex latter condition, I made the following transformation$$\begin{align} &x=-a+b+c\\ &y=a-b+c\\ &z=a+b-c \end{align}$$so that $x+y+z=a+b+c.$

Now,$$S=1\\ \Leftrightarrow xyz(x+y+z)=16\\ \mbox{which } x,y,z>0 \mbox{ and } x,y,z\in \mathbb{Q}$$ The equation is simple. However, I don’t know how to cope with the condition that $x\, y\,z$ are all positive rational numbers.It seems hard to deform.

I have known that $(\frac{3}{2},\frac{5}{3},\frac{17}{6})$ is a solution to the problem and there are more. But I don’t know how these numbers are calculated. I am puzzled by these figures.

Sint
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  • There's an answer here, but I don't know how it was found: https://math.stackexchange.com/a/2398172/87023 – Chris Culter Mar 04 '18 at 06:46
  • I think we're bumping our heads up against irrationality here. Firstly, any rational triangle can have its sides multiplied by the GCD of the denominators of the sides to make it an integral triangle. So, we're asking "is there an integral triangle with a rational area?" I'm thinking "no," and would like to come up with a synthetic answer to that (just because i'm old school). – estragon Mar 05 '18 at 05:11
  • hmmm... 9,10,17 works, so they're out there. Off to Heron's Theorem. – estragon Mar 05 '18 at 18:31
  • @estragon Thanks for your comment. Your thought gives a new way to deal with the problem. It avoids rational number and changes it to integers. I am trying this way but find another problem: if the sides of the triangle are multiplied by a number to have it integral, the area will be a SQUARED number. I find it hard to do with squared numbers. I am still stuck. – Sint Mar 06 '18 at 03:21

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