Can we have a theorem that makes Pythagoras numbers without any error? #proof

Question

Can we have a theorem that makes Pythagora's numbers without any errors?

Answer 1

I assume that by "Pythagorean number" you mean "Pythagorean triple" - that is, you're looking at generating triples $(a,b,c)$ such that $a^2+b^2=c^2$.

The answer is yes. In fact, the situation is as good as possible: we have a formula which enumerates only and all of the Pythagorean triples, and a proof that it works.

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First, the easy part - generating lots of triples:

For $0<n<m$, the triple $$a=m^2-n^2,\quad b=2mn, \quad c=m^2+n^2$$ is a Pythagorean triple.

This is easy to prove: $$a^2+b^2=(m^4-2m^2n^2+n^4)+4m^2n^2=m^4+2m^2n^2+n^4=(m^2+n^2)^2=c^2.$$ So immediately this gives us a way to generate infinitely many Pythagorean triples easily.

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It takes a bit more work to get every Pythagorean triple. First, note that the method above doesn't get all triples - e.g. it won't produce $a=9, b=12, c=15$ (check this!). We can fix this by throwing on another parameter:

Given $0<n<m$ and arbitrary $k>0$, the triple $$a=k(m^2-n^2),\quad b=k(2mn), \quad c=k(m^2+n^2)$$ is a Pythagorean triple; moreover, every Pythagorean triple has this form.

The proof that this always gives a Pythagorean triple is identical to the previous argument (just factor away the $k$s); the proof that every triple is gotten this way (which really just boils down to showing that every primitive triple was gotten the original way) takes more work - and in particular isn't just a direct algebraic simplification - but is still elementary.

Answer 2

Pick any odd number “A”. “B” = (AA+1)/2, “C”=B-1.

A,B and C are Pythagorean numbers.