More Squares in Pythag Triples

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Let a,b,ca,b,c be a Pythagorean Triple where a<b<ca < b < c.

Problem 1: Find the condition that a,b,ca,b,c need to satisfy such that b+cb+c is a perfect square.

Problem 2: Find the condition that a,b,ca,b,c need to satisfy such that a+ca+c is a perfect square.

Problem 3: If you can, find a condition for a,b,ca,b,c for a+ba+b to be a perfect square.

Note by Daniel Liu
5 years, 2 months ago

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Considering a, b, and c natural numbers:

For problem 1, a should be odd. For problem 2, a should be even. Can't figure out problem 3, as it seems impossible for natural numbers.

Niket Jain - 5 years, 2 months ago

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(9,40,41) satisfies problem 3's conditions.

Tan Li Xuan - 5 years, 2 months ago

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The general Pythagorean triple solutions are k.2ab,k(a2b2)k.2ab, k(a^2-b^2) and k(a2+b2)k(a^2+b^2).For #2 it suffices that k is a perfect square.For #1 2k needs to be a perfect square.

Bogdan Simeonov - 5 years, 2 months ago

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But we require a<b<c a < b < c . It is not true that 2xy<(x2y2)<(x2+y2) 2xy < ( x^2 - y^2) < (x^2 + y^2 ) (where I changed your notation from ab to xy).

Calvin Lin Staff - 5 years, 2 months ago

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But every pair of numbers is in the problems, so that doesn't change it very much.

Bogdan Simeonov - 5 years, 2 months ago

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I'm not really doing any proofs. Here's my answer for #1. I'll do the rest when I have more time. Great note though! :D

Problem 1: If bb and cc are consecutive integers that add to a perfect odd square. For example, the general formula for an odd number n=2x+1n=2x+1 for x>1x>1 is

(2x+1,(2x+1)22,(2x+1)22)(2x+1, \lfloor\dfrac{(2x+1)^2}{2}\rfloor, \lceil\dfrac{(2x+1)^2}{2}\rceil)

Finn Hulse - 5 years, 2 months ago

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What about (8,15,17)?

Daniel Liu - 5 years, 2 months ago

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What about it?

Finn Hulse - 5 years, 2 months ago

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@Finn Hulse It doesn't satisfy your requirement but it still satisfies #1...

Daniel Liu - 5 years, 2 months ago

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@Daniel Liu But 32 isn't a perfect square.

Tan Li Xuan - 5 years, 2 months ago

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@Tan Li Xuan Yeah...

Finn Hulse - 5 years, 2 months ago

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@Finn Hulse but 17+817+8 is

Daniel Remo - 5 years, 2 months ago

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@Daniel Remo True, but we're talking about Problem 1, where b+cb+c is a perfect square.

Finn Hulse - 5 years, 2 months ago

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@Finn Hulse Oh oops, I accidentally thought you were talking about #2, don't know what happened. But I can still find other triplets that have b+cb+c be square and not satisfy your requirement I thnik.

Daniel Liu - 5 years, 2 months ago

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@Daniel Liu Yes I "thnik" you can. :D

Finn Hulse - 5 years, 2 months ago

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