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.

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## Comments

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a,b, andcnatural numbers:For

problem 1,ashould be odd. Forproblem 2,ashould be even. Can't figure outproblem 3, as it seems impossible for natural numbers.Log in to reply

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

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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.

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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).

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

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

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What about (8,15,17)?

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What about it?

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