an interesting result regarding Pythagorean triples...

Рет қаралды 19,354

Күн бұрын

Пікірлер: 39

@Bodyknock 11 ай бұрын

Funnily enough I thought the Lemma that a+b+c always divides ab in a Pythagorean Triple was the cooler result. It's just a really neat looking simple relationship between a, b, and c. 🙂

@DeanCalhoun 11 ай бұрын

I agree!

@miraj2264 11 ай бұрын

At 11:00, he's alluding to the case written out on the board. If a number x is odd, you can write it as 2n+1 for some n. So x^2 = 4n^2 + 4n + 1. If we take this mod 4, we see that an odd number squared mod 4 is always 1. Similarly, if y is even, you can write it as 2m for some m. So y^2 = 4m^2. If we take this mod 4, we get 0. We're dealing with Pythagorean triples here so we have a^2 + b^2 = c^2. If a and b are odd, that means the LHS is 2 mod 4. But the RHS can only be 1 if c is odd or 0 if c is even. Thus, a and b can't both be odd. So the only possible Pythagorean triples would be 2 evens (like 6^2 + 8^2 = 10^2) or an odd and an even (like 3^2 + 4^2 = 5^2).

@baerlauchstal 11 ай бұрын

Yes, the parametrization method works I reckon. Let a = 2pq, b = p^2 - q^2, c = p^2 + q^2. Assume the triple is primitive. (Easy to show that if the claim holds for primitive triples it holds for all triples; and the parametrization doesn't necessarily work in any case if the triple isn't primitive.) Then p and q are coprime (because gcd(a,b,c) is obviously divisible by gcd(p,q)). It follows that p and (p + q) are coprime. Also p + q is odd; otherwise, each of a, b and c is even, contradicting the hypothesis of primitivity. Hence 2p and (p + q) are coprime. Finally, easy algebra gives a + b + c = 2p(p + q). Now, consider S = a^r + b^r + c^r = (2pq)^r + (p^2-q^2)^r + (p^2+q^2)^r, where r is odd. S is divisible by 2p, because (a) this is clearly true of (2pq)^r, and (b) the other two terms, when expanded, sum to 2p^(2r) + 2 (r choose 2) p^(2r-4)q^4 + ...; the power of p in each term is twice a positive odd number, and therefore greater than zero; hence 2p divides each term. And S is congruent modulo (p + q) to (2(-q)q)^r + (q^2 - q^2)^r + (q^2 + q^2)^r = -2^r q^(2r) + 2^r q^(2r) = 0. Hence S is divisible by both 2p and (p + q), and since these are coprime, by 2p(p + q). And we're done. This might be tidyable-up.

@baerlauchstal 11 ай бұрын

Here's a tidying up of one bit. To show that S is divisible by 2p, we observe that this is true of (2pq)^r as before, and then show that the sum of the second and third terms is divisible by 2p^2. That's true because that sum is congruent, modulo 2p^2, to (-p^2-q^2)^r + (p^2+q^2)^r = -(p^2 + q^2)^r + (p^2 + q^2)^r = 0. Slightly neater?

@ivonovicsklrtweros 11 ай бұрын

where is the backflip :(

@MathsMadeSimple101 11 ай бұрын

I'm new here, what back flip?

@ezequielangelucci1263 11 ай бұрын

@@MathsMadeSimple101 He used to do backflips in the videos.

@ivonovicsklrtweros 11 ай бұрын

@@MathsMadeSimple101 ı have watched all videos of this super amazing teacher, This guy used to always do backflips in his videos :(

@chrisbhag 11 ай бұрын

Thanks for sharing professor! I Always enjoy your work! Wishing you all the best ❤🎉

@goodplacetostop2973 11 ай бұрын

17:19

@gregsarnecki7581 11 ай бұрын

‘Proof by hope’ - love it! It gives hope to all of us that hard maths can be overcome.

@pascalochem4256 11 ай бұрын

At 11:00, start with a^2+b^2=c^2, so a^2+b^2=c^2 (mod 2), so a+b=c (mod 2) since x^2=x (mod 2), so a+b-c=0 (mod 2), so a+b-c is even.

@DrR0BERT 11 ай бұрын

I tackled this using the parameterization. It was tedious, and if I can streamline it I will. A few things I found is that a+b+c reduces to 2m^2+2mn = 2m (m+n). 2m and (m+n) are relatively prime. It's easy to show that 2m divides a^(2n+1)+b^(2n+1)+c^(2n+1). It gets tougher to show that m+n does as well.

@aadfg0 11 ай бұрын

This was my approach. Why is it tough? Let q = m+n. a = 2nm = -2n^2 mod q, b = 0 mod q, c = m^2+n^2 = 2n^2 mod q so the sum is (2n^2)^(2k+1) + (-2n^2)^(2k+1) = 0 mod q.

@DrR0BERT 11 ай бұрын

@@aadfg0 Oh! I took the long way around.

@yossigil739 11 ай бұрын

Can I suggest two problems for a video on variations Pythagorean triples? The natural question: What’s the parametrized form of a Pythagorean quartet? A variation is find all non trivial a,b,c,d\in\mathbf N such that a^2 + b^2 = c^2 + d^2, =>c^2-a^2 = d^2 - b^2 => (c-a)(c+a) = (d - b) (d +b). Denote x = c -a, y = c + a, z = d - b, u = d - b, v = d + b. => u v = xy , denote n = u v. Then an algorithm for generating all solutions is itertiating over all non primes and generating all distinct permutations of it. Not sure there is a parametrization though.

@prathikkannan3324 11 ай бұрын

Satisfying. This was a great problem solving solution.

@AnkhArcRod 11 ай бұрын

At 5:27, though it is obvious that (a+b+c) cannot | a+b, it is not entirely obvious why gcd(a+b, a+b+c) = 1 (which, in general, is not true; eg. (6,8,10)). Thus, it would have been appropriate to write -ab(a+b) as -ab(a+b+c-c) = -ab(a+b+c) + abc. Thus, (a+b+c) should | abc. It turns out that indeed a+b+c |ab, but it would be wrong to jump to that conclusion at the step where we did.

@TheEternalVortex42 11 ай бұрын

It doesn't really matter since a+b+c|ab => a+b+c+|ab(a+b). So you're right that a more complex condition could have been required but it turns out it wasn't.

@Qermaq 11 ай бұрын

Nice. I thought about using the parameterization method you mentioned, but I doubt it would be a better proof. Would seem to me it would mostly parallel what you did here just with a bit more of a mess.

@bourhinorc1421 11 ай бұрын

Didn't you have a video "sum(a_k^n) is multiple of sum(a_k)?" Do someone have the link to it?

@vcvartak7111 11 ай бұрын

First time i understood after watching so many problems

@Timmmmartin 11 ай бұрын

Without proving it, I'm pretty sure that by replacing b with -b in both the numerator and the denominator, another very similar result occurs, where b>a.

@師太滅絕 11 ай бұрын

I always have a question, take 3,4,5 the triplets as example, can 3 or 4 be in another triplets? (Of course not 5, which can be in 5, 12, 13 triplets. Essentially, can the two smaller numbers of a P-triplets occur in another triplets (as the two smaller numbers) thanks

@stefanalecu9532 11 ай бұрын

There are no Pythagorean triangles in which the hypotenuse and one leg are the legs of another Pythagorean triangle (this is a corollary of Fermat's right triangle theorem), so that possibility is ruled out. You can have at most either that leg or the hypothenuse to be the leg of another triangle. You can find examples of distinct (a, b) pairs that satisfy the Pythagorean property and sum to the same c². If you have a primitive Pythagorean triple (where the gcd(a,b)=1) then you can generate it using two positive integers m>n>0 with m+n odd and gcd(m,n)=1, as Euclid has shown: (m²-n², 2mn, m²+n²). The hypothenuse is necessarily odd, so if you have (p²-q², 2pq, p²+q²) then it necessarily follows that you have m²+n²=p²-q² OR m²+n²=p²+q² (in which case there are an infinite number of triples with the same c²).

@yoav613 11 ай бұрын

Very nice

@davidgillies620 11 ай бұрын

If a, b, c are a primitive Pythagorean triple, one of a, b must be even and the other odd (if both odd c^2 = 2 mod 4, impossible; if both even a, b, c not primitive).

@maxxis4035 11 ай бұрын

I have a question for anyone really. So I have to determine if the following integral converge or diverge: from 0 to (pi)^2 of (1/[1- cos(sqrt(x)]), any help is appriciated.

@damadclown 11 ай бұрын

Usually in these cases you can use a taylor series expansion around 0 (it's the only point where the integral can diverge): you'll get something like f(x) ~ 2/x which means the integral diverges. Hope it helps

@maxxis4035 11 ай бұрын

@@damadclown Ill look into it, thanks!

@baerlauchstal 11 ай бұрын

@@maxxis4035 It's easier than that I think. Can you show that the integrand is bounded below by 2/x?

@divisix024 11 ай бұрын

Remark: The identity a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc, which holds for all a,b,c without the need for Pythagorean triple, also necessitates the lemma.

@ismailozdas1894 11 ай бұрын

this theorem without pythagorean it will not be true because dividing 1^3 + 1^3 + 2^3 /1+1+2 is not integer

@divisix024 11 ай бұрын

@@ismailozdas1894 Indeed, the lemma needs a,b,c to be a Pythagorean triple, only the identity does not. Notice how that definitely doesn’t imply a^3+b^3+c^3 is divisible by a+b+c.

@enpeacemusic192 11 ай бұрын

Would be cool to derive an interesting identity that holds if and only if it’s an Pythagorean triple (that isn’t just the standard identity), since this one obviously is satisfied triples that aren’t Pythagorean (the identity is symmetric for all permutations of a, b, c while Pythagorean triples are symmetric for permutations of a and b)