Prove that every product of Pythagorean triples is a multiple of 60 (ILIEKMATHPHYSICS)

[@jesusthroughmary]

So 3-4-5 truly is the primordial Pythagorean triple

[@darkrozen4110]

Amazing work, modular arithmetic in proofs are always interesting to see.

[@The_lenny]

Amazing result

[@davidbrisbane7206]

Euclid's formula for generating Pythagorean triples given an arbitrary pair of integers m and n with m > n > 0 is: a = m² - n², b = 2mn, c = m² + n². So, abc = 2mn(m² - n²)(m² + n²) = 2mn(m⁴ - n⁴) Clearly, 4 | 2mn, if either m or n, or both are even. So, suppose by way of contradictions that m and n are both odd. Then m⁴ - n⁴ is even, so 4 | 2mn(m⁴ - n⁴). If 3 | n (or m) and 5 | n (or m), then we are done, as then 3*4*5 = 60 | 2mn(m⁴ - n⁴) So now suppose that 3 doesn't divide m or n, and that 5 doesn't divide m or n. If 3 doesn't divide n, then n ≣ 1 or 2 (mod 3) So, n⁴ ≣ 1 (mod 3) in both cases. If 3 doesn't divide m, then m ≣ 1 or 2 (mod 3) So, m⁴ ≣ 1 (mod 3) in both cases. So, m⁴ - n⁴ ≣ 1 - 1 ≣ 0 (mod 3) So, 3 | 2mn(m⁴ - n⁴). If 5 doesn't divide n, then n ≣ 1, 2, 3 or 4 (mod 5) So, n⁴ ≣ 1 (mod 5) in all cases. If 5 doesn't divide m, then m ≣ 1, 2, 3 or 4 (mod 5) So, m⁴ ≣ 1 (mod 5) in all cases. So, m⁴ - n⁴ ≣ 1 - 1 ≣ 0 (mod 5) So, 5 | 2mn(m⁴ - n⁴). Putting it all together, 3 | 2mn(m⁴ - n⁴) and 4 | 2mn(m⁴ - n⁴) and 5 | 2mn(m⁴ - n⁴ So, 60 | abc.

[@dogbiscuituk]

Yes! So starting from abc = mnp, where p = m⁴ - n⁴, for any base 2 ≤ d ≤ 5, if neither a nor b is 0 (mod d), then they are both 1 (mod d), so m⁴ - n⁴ is 0 (mod d), and abc is 0 (mod d), QED! I know this doesn't add much to your presentation or to the video, just wanted to highlight an opportunity to link this to a discussion of quadratic / quartic residues and reciprocities, which could be fun.

[@user-ou1yw4cg6y]

This is so amazing

[@davidemasi__]

Great approach as always 💪🏼

[@orangeinks6681]

Wow! This is an amazing video, enjoyed it throughout!

[@lumbersnackenterprises]

It would make sense since Pythagoras learned math from the Chaldeans who used a base 60 system. The reason I bring this up is because their math revolves around the circle, which in my opinion is just a collection of infinite Triangles

[@justusschoenmakers8987]

Pythagoras doesnt have anything to do with this. Aliens could come up with exactly the same formula. Its only the nature of base 10 that gives this interesting result

[@lumbersnackenterprises]

@justusschoenmakers8987 Pythagoras is the largest contributor to Western Mathematics and our Base 10 system is a hybrid of the Chaldean Base 60 system, the Egyptian Base 10 system, and the Indian Base 10 Symbol Logic. It's less about Pythagoras and more about the innateness of the Base 60 system that was created with Circles and Triangles in mind. The fact that the product of any given Pythagorean Triple is a multiple of 60 is likely a direct result of this. The Sexigesimal system of Babylon was hyper advanced for its time because it was an evolution of Unary (Tally Marks) which basic counting is the Purest Form of Mathematics.

[@justusschoenmakers8987]

@@lumbersnackenterprises no a^2+b^2=c^2 is fundamental and arises in every number system. Counting in base 60 or in binary it doesnt change the formula.

[@lumbersnackenterprises]

@justusschoenmakers8987 First, you said Pythagoras doesn't have anything to do with this and then cited the Pythagorean Theorem. That's like saying Euler has nothing to do with exponential growth and decay despite the constant 2.718281828 being named after him. Second, obviously the constant doesn't change based on the base system and that was never my argument. It's stating a fact about the nature of a Base 60 system that is core to the current mathematical model is a relevant part of this core equation. Not only that, it's pretty obvious something like this would happen because the angles of a Perfect Triangle are all 60° so a perfected triangle equation such as the Pythagorean Triples would likely have something to do with multiples of 60. Third, What kind of argument even is stating that an alien could have come up with the equation anyway. It was Pythagoras who discovered and is credited with it and it's in his understanding of Mathematics that he learned from the Egyptians and Chaldeans that allows the logic for this guys proof to even be stated

[@justusschoenmakers8987]

@@lumbersnackenterprises clearly you dont know what youre talking about. Dive in the books lad!

[@gregevgeni1864]

Nice 🎉🎉

[@markgraham2312]

Very nice!

[@Anonymous-zp4hb]

Nice. I solved it a similar way... 0*0=0 (mod 3) 1*1=1 (mod 3) 2*2=4=1 (mod 3) So nn = 0 OR 1 (mod 3) 0+0=0 [yep] 0+1=1 [yep] 1+1=2 [nope] There's a zero on the left in all of the allowed cases So there's a multiple of 3 on the left 0*0=0 (mod 4) 1*1=1 (mod 4) 2*2=4=0 (mod 4) 3*3=9=1 (mod 4) So nn = 0 OR 1 (mod 4) 0+0=0 [yep] 0+1=1 [yep] 1+1=2 [nope] There's a zero on the left in all of the allowed cases So there's a multiple of 4 on the left 0*0=0 (mod 5) 1*1=1 (mod 5) 2*2=4 (mod 5) 3*3=9=4 (mod 5) 4*4=16=1 (mod 5) So nn = 0 OR 1 OR 4 (mod 5) 0+0=0 [yep] 0+1=1 [yep] 0+4=4 [yep] 1+1=2 [nope] 1+4=5=0 [yep] 4+4=8=3 [nope] There's a zero in all of the allowed cases So there's a multiple of 5 somewhere.

[@SurprisedDivingBoard-vu9rz]

Base into first 2 symbols. Base 10. 2 and 3 even odd mix.