Please help devise a formula that will solve x^2 ┴ (x ┴ 7)^2 = c^2 , where c is a whole number. Thank you.

- December 16th 2010, 01:14 AMribbielooking for a formula to solve a particular pythagorean triple
Please help devise a formula that will solve x^2 ┴ (x ┴ 7)^2 = c^2 , where c is a whole number. Thank you.

- December 16th 2010, 01:51 AMearboth
- December 16th 2010, 02:10 AMribbie
Yes!

Quote:

I've found (by brute force) 1, 7, 13, 17, 35, 103

Quote:

4. Probably there is an unlimited number of values which will satisfy the equation.

I was thinking we could use the rule

but I got lost trying to figure it out. - December 16th 2010, 06:30 AMOpalg
There is a page of information here about the infinite sequence of solutions of the Diophantine equation . Those solutions obviously satisfy the equation , where If you then put , you will have an infinite sequence of solutions of the equation

I think that in that way you will get all the solutions for which x is a multiple of 7. But there are also plenty of other solutions. - December 16th 2010, 08:27 AMWilmerHmmm....
- December 17th 2010, 03:16 AMribbie
I'm not sure I understand this. If then I can use , for example, but that gives me http://www.mathhelpforum.com/math-he...603a864af1.png, where but that means that c is not a whole integer (which is what I am trying to find.)

- December 17th 2010, 06:41 AMOpalg
No, you can't use because that would come from , which is not an allowable value of In fact, must be one of the Pell numbers Those numbers don't look as though they will be integers, but when you expand the binomial series for , the terms with all cancel out, leaving you with an integer. The first few such integers are When you form , you get the numbers In each case, you can check that is a square. As I said in my previous comment, you don't get anything like all the possible values of in that way. But I think that you do get all those that are multiples of 7.