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How to prove that all Heronian triangles can be found using formulas described here?

I understand that the described substitution will give Heronian triangle, but how to prove that using the described substitution will give all Heronian triangles (with scaling solutions, for example $(12,10,10)$ is obtained when multiplying $(6,5,5)$ by $2$)?

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Perhaps related:… – joriki Mar 1 '12 at 20:02
This appears to be the Carmichael publication referred to in the Wikipedia article. – Peter Phipps May 1 '12 at 10:57

Well, understanding that Heronian triangles are cyclic is a start. So the scaled up($\times n$, say) version of a Heronian would fit precisely into a circle of twice the diameter.

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What do you mean, "Heronian triangles are cyclic"? Every triangle, Heronian or not, can be inscribed in a circle. – Gerry Myerson Mar 1 '12 at 22:54

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