Pythagorean tripples where 2 of the sides differs by 1

[Lufen]

Didn't really know which category this problem would fit in but I hope this works. I have a homework assignment about pythagoras theorem. There are many examples of Pythagorean tripples where 2 of the sides differs by exactly 1. Examples of this could be (3, 4, 5) , (5, 12, 13) or (20, 21, 29) and I am asked to find a method that can easly find all of these. I started of by assuming that these tripples in general could either be written as (a, a +1, b) or (a, b, b+1). In the first case I assumed that (a² + (a+1)² = b²) and then that (2a²+2a+1 = b²) and then took square root of both sides to just get b alone but this doesnt really this doesnt really seem as a good solution at all. I dont really know what I could do else.

 

[pka]

Didn't really know which category this problem would fit in but I hope this works. I have a homework assignment about pythagoras theorem. There are many examples of Pythagorean tripples where 2 of the sides differs by exactly 1. Examples of this could be (3, 4, 5) , (5, 12, 13) or (20, 21, 29) and I am asked to find a method that can easly find all of these. I started of by assuming that these tripples in general could either be written as (a, a +1, b) or (a, b, b+1). In the first case I assumed that (a² + (a+1)² = b²) and then that (2a²+2a+1 = b²)

If you are at a school with a good library, look for An Introduction the History of Mathematics by Howard Eves.
Here is some material from that text. If \(m\) is an odd natural number then \(m^2+\left(\dfrac{m^2-1}{2}\right)^2=\left(\dfrac{m^2+1}{2}\right)^2\) form a set of Pythagorean triples. Ex \(m=7\) gives \(7, 24, 25\) Note two are consecutive.
Mover, for any natural number greater than zero \((2n)^2+\left(n^2-1\right)^2=\left(n^2-1\right)^2\) also form such a set. Attributed to Plato (ca. 380 BCE)

 

[Dr.Peterson]

Didn't really know which category this problem would fit in but I hope this works. I have a homework assignment about pythagoras theorem. There are many examples of Pythagorean tripples where 2 of the sides differs by exactly 1. Examples of this could be (3, 4, 5) , (5, 12, 13) or (20, 21, 29) and I am asked to find a method that can easily find all of these. I started of by assuming that these tripples in general could either be written as (a, a +1, b) or (a, b, b+1). In the first case I assumed that (a² + (a+1)² = b²) and then that (2a²+2a+1 = b²) and then took square root of both sides to just get b alone but this doesnt really this doesnt really seem as a good solution at all. I dont really know what I could do else.

Have you learned the formula for Pythagorean triples? Have you tried determining what values of the parameters yield two consecutive numbers?