Wha is the proof that there are infinitely many right angled (non-similar) triangles whose sides have integral lengths? I know that this is equivalent to showing that there are infinite pythagorean triples, which can be proven easily, but I would like to know about any purely geometrical proof.
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(3, 4, 5) is a Pythagorean triple. So is K(3, 4, 5) where K is a positive integer. There are infinite many already. – Mick May 27 '14 at 14:31
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1I meant a geometrical proof, not one which uses infinitude of pythagorean triples. – A Bajaj May 27 '14 at 14:34
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You might want to put the geometrical aspect of this in the title - as it is, it looks like a run of the mill question. – Peter Woolfitt May 27 '14 at 14:38
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@Mick, I had the same idea but OP asks for non-similar triangles. – vadim123 May 27 '14 at 14:41
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@Peter Woolfitt Is "geometric proof of infitude of right triangles with integral sides" better? – A Bajaj May 27 '14 at 14:42
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@ABajaj Yeah, I think it is. – Peter Woolfitt May 27 '14 at 14:45
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The work of Roger Vogeler might be of interest here, consider my brief writeup here – abiessu May 27 '14 at 14:54
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Interesting, but I was thinking of a proof along the lines of synthetic geometry (is that the right term?) – A Bajaj May 27 '14 at 15:02
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"Integral" is not a purely geometric notion, so there cannot be a purely geometric proof. – May 27 '14 at 16:40
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@vadim123 You are right. I have overlooked the non-similar part. – Mick May 28 '14 at 03:50
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Consider the following diagram:
So the area of the green part must be equal to the area of the orange part by the pythagorean theorem. However, the area of the green part is $2X+1$. Hence, we can make the area of the orange part any odd number with a positive integer choice of $X$. In particular we can look at the sequence where $2X+1=p_i^2$ where $p_i$ is the $i$-th odd prime. Hence we have an infinite sequence of right triangles with of side lengths $p_i,X,X+1$ where $2X+1=p_i^2$. Note that none of these triangles are similar because one sidelength of each is prime.
Peter Woolfitt
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Yeah, I should point out though that it is not quite purely geometric because it inherently relies on the infinitude of primes - which is an algebraic fact. – Peter Woolfitt May 27 '14 at 16:46
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But Euclid gave a geometric proof for infinitude of primes, didn't he? Then can we say that this is a purely geometric proof? – A Bajaj May 27 '14 at 16:50
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Whoops, I see I should have said "which I only know as an algebraic fact." I don't know a geometric proof for the infinitude of primes, but you're right, I would guess that one exists, and if it does, this could be just a geometric argument. – Peter Woolfitt May 27 '14 at 16:58
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See Euclid,Book 9, prop. 20 http://aleph0.clarku.edu/~djoyce/java/elements/bookIX/propIX20.html – A Bajaj May 27 '14 at 17:00
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Nb The proof is geometric only in the sense that for Euclid, there is no notion of numbers; when he says numbet, he means a length of a given line. Using these concepts he proves surprisingly , many results in number theory. – A Bajaj May 27 '14 at 17:10
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The wikipedia page on Pythagorean triples describes the geometric proof of the enumeration of Pythagorean triples. In brief, there are two steps. First, there is a one-to-one correspondence between primitive Pythagorean triples $(a,b,c)$ and rational points $(r,s)=(a/c,b/c)$ on the unit circle $S^1$. Second, sterographic projection, radiating away from the north pole $N$ and projecting $S^1 - N$ to the $x$-axis, defines a one-to-one correspondence $S^1 - N \leftrightarrow \mathbb{R}$. Furthermore, stereographic projection restricts to a one-to-one correspondence between rational points $(r,s)$ on the unit circle and the set $\mathbb{Q}$ of rational numbers on the $x$-axis (this is where you need some formulas, which can be found in the wikipedia link provided).
Here are the formulas for stereographic projection, which I am copying from the wikipedia page. If $P = (r,s)$ is a rational point on the unit circle and if $P' = (m/n,0)$ is a rational point on the real line, so that stereographic projection relates $P$ to $P'$, then the formulas relating the coordinates of $P$ and of $P'$, are: $$r = \frac{2mn}{m^2+n^2}, \,\,\, s = \frac{m^2-n^2}{m^2+n^2} $$ and $$\frac{m}{n} = \frac{r}{1-s} $$ These formulas can be derived using elementary geometry, starting from the picture of stereographic projection.
Lee Mosher
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I don't really understand much after the second sentence. Could you please explain more simply? Thank you. – A Bajaj May 27 '14 at 15:57
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I rewrote a little, and added a link to the picture of stereographic projection that can be found in wikipedia link provided. – Lee Mosher May 27 '14 at 16:29
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I haven't learn about projections yet. Is this usually taught in high schools? – A Bajaj May 27 '14 at 16:33
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I don't think that projective geometry is usually taught in high school. But I would guess that if you have a good teacher then they might recommend a book on projective geometry at the right level for you. – Lee Mosher May 27 '14 at 16:48
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My teacher is very good, but doesn't have a maths degree... could you perhaps tell me of a good book on the subject? I know euclidean geometry upto book 3 (and most of book 6), elementary algebra ; and combinatorics (very basic) and set theory, and a bit of calculus. – A Bajaj May 27 '14 at 16:54
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I added the actual formulas for stereographic projection. If you use the picture that I linked, you should be able to derive those formulas. Then that would be all that you need to understand Pythagorean triples. – Lee Mosher May 27 '14 at 16:56