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# More Squares in Pythag Triples

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

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

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

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

Note by Daniel Liu
2 years, 4 months ago

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The general Pythagorean triple solutions are $$k.2ab, k(a^2-b^2)$$ and $$k(a^2+b^2)$$.For #2 it suffices that k is a perfect square.For #1 2k needs to be a perfect square. · 2 years, 4 months ago

But we require $$a < b < c$$. It is not true that $$2xy < ( x^2 - y^2) < (x^2 + y^2 )$$ (where I changed your notation from ab to xy). Staff · 2 years, 4 months ago

But every pair of numbers is in the problems, so that doesn't change it very much. · 2 years, 4 months ago

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. · 2 years, 4 months ago

(9,40,41) satisfies problem 3's conditions. · 2 years, 4 months ago

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 $$b$$ and $$c$$ are consecutive integers that add to a perfect odd square. For example, the general formula for an odd number $$n=2x+1$$ for $$x>1$$ is

$(2x+1, \lfloor\dfrac{(2x+1)^2}{2}\rfloor, \lceil\dfrac{(2x+1)^2}{2}\rceil)$ · 2 years, 4 months ago

What about (8,15,17)? · 2 years, 4 months ago

What about it? · 2 years, 4 months ago

It doesn't satisfy your requirement but it still satisfies #1... · 2 years, 4 months ago

But 32 isn't a perfect square. · 2 years, 4 months ago

Yeah... · 2 years, 4 months ago

but $$17+8$$ is · 2 years, 4 months ago

True, but we're talking about Problem 1, where $$b+c$$ is a perfect square. · 2 years, 4 months ago

Oh oops, I accidentally thought you were talking about #2, don't know what happened. But I can still find other triplets that have $$b+c$$ be square and not satisfy your requirement I thnik. · 2 years, 4 months ago