## Saturday, January 24, 2009

### Finding Pythagorean triples with Pell's equation

An earlier post I made makes mention of relating Pell's equation to Pythagorean triples, and it turns out that you can just solve for Pythagorean triples using solutions from certain Pell's equations:

Given x2 - Dy2 = 1, and x = 1 mod D or x = -1 mod D, from the prior post I reference above I have that

(D-1)j2 + (j+1)2 = (x+y)2

or

(D-1)j2 + (j-1)2 = (x+y)2

where

j = ((x+Dy)-1)/D or j = ((x+Dy)+1)/D

which gives a Pythagorean triple whenever D-1 is a square, for instance with D=2, I have a solution with x=17, y=12, as 172 - 2(12)2 = 1, so

j = ((17+2(12))-1)/2 = 20 is a solution giving:

202 + 212 = 292

Notice that different types of Pythagorean triples are included as D increases in size, as for instance, with D=5, you have triples of the form:

(2j)2 + (j+/-1)2 = (x+y)2

(where I use "+/-" as an OR as above as it's less tedious)

so given a solution to x2 - 5y2 = 1, you have solutions.

e.g. x=9, y = 4, works as 92 - 5(4)2 = 1, so j = (9 + 5(4) + 1)/5 = 6 is a solution. And then

122 + 52 = 132.

Notice too that when D-1 is not a square you get integer solutions to ellipses, and you also get the result that with a natural number n, there are always an infinity of integer, non-zero solutions to the equation: nu2 + v2 = w2

The question now arises, can you go in the other direction and solve for a Pell's equation using a Pythagorean triple?

The answer is, yes. Consider, 242 + 72 = 252.

If 7 is j+/-1, I can try j = 6 or 8, and find D=17 or D=10.

With j=6, D=17, I have j = ((x+Dy) -/+1)/D, so

17(6) = x+17y -/+ 1, so x+17y = 17(6)+/-1, and x+y = 25, so

16y = 17(6)+/-1 - 25, which doesn't give an integer y.

But trying j=8, D=10:

x+10y = 10(8)+/-1, so 9y = 10(8)+/-1 - 25, gives y = 6 as a solution, so x = 19, and

192 - 10(6)2 = 1.

So you can go in either direction.

Which then proves an infinite number of solutions to the more general form of Pell's Equation.

James Harris

## Sunday, January 11, 2009

### To discuss go to my math group

I am increasingly interested in the resources that My Math Group provides, as reality is that it's extremely difficult to get a comment on this blog itself, and besides, the format at the group is more naturally suited for discussion anyway.

So if you've wanted to talk about my research you have the opportunity available.

James Harris