"If $\lambda, m,$ and $n$ are positive rational numbers, and $m > n$, then $\lambda(m^2 − n^2), 2\lambda mn$, and $\lambda(m^2 + n^2)$ are positive rational numbers. Hence show how to determine any number of right-angled triangles the lengths of all of whose sides are rational."

What does he mean by "how to determine any number of right angled triangles"?

  • 4
    $\begingroup$ You’re being asked to show how to construct infinitely many different right triangles with rational sides. $\endgroup$ – Brian M. Scott May 17 '13 at 19:57
  • $\begingroup$ @BrianM.Scott different=non-similar? $\endgroup$ – Federica Maggioni May 17 '13 at 20:07
  • $\begingroup$ It doesn’t actually say so, but you can, so you might as well: that’s clearly a nicer solution. $\endgroup$ – Brian M. Scott May 17 '13 at 20:08
  • $\begingroup$ I think different here means non-congruent. $\endgroup$ – Qiaochu Yuan May 17 '13 at 20:22
  • $\begingroup$ @FedericaMaggioni i'm trying to understand the problem myself , if you have an idea it would be great if you could give me a hint or two. I was trying to bring the expressions in quadratic form.. $\endgroup$ – sigmatau May 17 '13 at 20:26

We know that $3$ lengths can form a right angled triangle iff those lengths satisfy $a^2+b^2=c^2$.

But no matter what we choose for $m, \; n$ and $ \lambda$. $$(2\lambda mn)^2+(\lambda(m^2-n^2))^2=4\lambda^2m^2n^2+\lambda^2m^4-2 \lambda^2m^2n^2+\lambda^2n^4$$ $$=\lambda^2m^4+2\lambda^2m^2n^2+\lambda^2n^4=(\lambda(m^2+n^2))^2$$ as we wanted.

So we can choose any number of rationals $m, \; n$ and $\lambda$ to produce any number of right-angled triangles.

A simple example of an infinite solution set is if you keep $m$ and $n$ constant, but vary $\lambda$, you will get infinite solutions as there are an infinite amount of rational numbers.


If $a,b,c$ are rational with $a\ne 0\ne b$ and $a^2+b^2=c^2$ then for some $n\in N,$ the numbers $|a n|,|b n|, |c n|$ are all in $N,$ so there exist $p,q,r\in N$ with $p\ne q$ and $\{|a n|,|b n|\}=\{r(p^2-q^2), 2 r p q\}$ and $|cn|=r(p^2+q^2),$ giving $\{a,b\}=\{(r/n)(p^2-q^2),(r/n)2 p q)\}$ and $|c|=(r/n)(p^2+q^2).$

Conversely, if $s\in Q^+$ and $p,q\in N$ with $p\ne q$ then $[s(p^2-q^2)]^2+[2 s p q]^2=[s(p^2+q^2)]^2.$

Remark: Long before Pythagoras,the Babylonians compiled long tables of natural numbers $a,b,c$ satisfying $a^2+b^2=c^2.$ It is unknown whether they had a formal proof of the theorem of Pythagoras or of the formula for Pythagorean triples. But they were good at arithmetic, with a number-base $60$ (sixty!) with "decimal" point and used "decimal" fractions (in base $60.$)


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