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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TopNewestThe 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. – Bogdan Simeonov · 3 years, 1 month ago

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– Calvin Lin Staff · 3 years, 1 month 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).Log in to reply

– Bogdan Simeonov · 3 years, 1 month ago

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

Considering

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. – Niket Jain · 3 years, 1 month agoLog in to reply

– Tan Li Xuan · 3 years, 1 month ago

(9,40,41) satisfies problem 3's conditions.Log in to reply

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\[(2x+1, \lfloor\dfrac{(2x+1)^2}{2}\rfloor, \lceil\dfrac{(2x+1)^2}{2}\rceil)\] – Finn Hulse · 3 years, 1 month ago

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– Daniel Liu · 3 years, 1 month ago

What about (8,15,17)?Log in to reply

– Finn Hulse · 3 years, 1 month ago

What about it?Log in to reply

– Daniel Liu · 3 years, 1 month ago

It doesn't satisfy your requirement but it still satisfies #1...Log in to reply

– Tan Li Xuan · 3 years, 1 month ago

But 32 isn't a perfect square.Log in to reply

– Finn Hulse · 3 years, 1 month ago

Yeah...Log in to reply

– Daniel Remo · 3 years, 1 month ago

but \(17+8\) isLog in to reply

– Finn Hulse · 3 years, 1 month ago

True, but we're talking about Problem 1, where \(b+c\) is a perfect square.Log in to reply

– Daniel Liu · 3 years, 1 month 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.Log in to reply

– Finn Hulse · 3 years, 1 month ago

Yes I "thnik" you can. :DLog in to reply