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.

#Algebra #PythagoreanTheorem #Square #PythagoreanTriples #Condition

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Comments

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 - 6 years, 12 months ago

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

Tan Li Xuan - 6 years, 12 months ago

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 - 6 years, 12 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).

Calvin Lin Staff - 6 years, 12 months ago

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

Bogdan Simeonov - 6 years, 12 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

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