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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, 7 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. Bogdan Simeonov · 2 years, 7 months ago

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@Bogdan Simeonov 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). Calvin Lin Staff · 2 years, 7 months ago

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@Calvin Lin But every pair of numbers is in the problems, so that doesn't change it very much. Bogdan Simeonov · 2 years, 7 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 · 2 years, 7 months ago

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@Niket Jain (9,40,41) satisfies problem 3's conditions. Tan Li Xuan · 2 years, 7 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 \(b\) 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 · 2 years, 7 months ago

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@Finn Hulse What about (8,15,17)? Daniel Liu · 2 years, 7 months ago

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@Daniel Liu What about it? Finn Hulse · 2 years, 7 months ago

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@Finn Hulse It doesn't satisfy your requirement but it still satisfies #1... Daniel Liu · 2 years, 7 months ago

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@Daniel Liu But 32 isn't a perfect square. Tan Li Xuan · 2 years, 7 months ago

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@Tan Li Xuan Yeah... Finn Hulse · 2 years, 7 months ago

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@Finn Hulse but \(17+8\) is Daniel Remo · 2 years, 7 months ago

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@Daniel Remo True, but we're talking about Problem 1, where \(b+c\) is a perfect square. Finn Hulse · 2 years, 7 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 \(b+c\) be square and not satisfy your requirement I thnik. Daniel Liu · 2 years, 7 months ago

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@Daniel Liu Yes I "thnik" you can. :D Finn Hulse · 2 years, 7 months ago

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