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

- Dec 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.

- Dec 16th 2010, 01:51 AMearboth
1. Do you mean

$\displaystyle x^2+(x+7)^2=c^2$

2. If so solve the equation for x:

$\displaystyle x^2+(x+7)^2-c^2=0$

3. To find those numbers c the discriminant must be a square. I've found (by brute force) 1, 7, 13, 17, 35, 103

4. Probably there is an unlimited number of values which will satisfy the equation. - Dec 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 $\displaystyle a = n^2 - m^2, b = 2nm, c = n^2 + m^2 $

but I got lost trying to figure it out. - Dec 16th 2010, 06:30 AMOpalg
There is a page of information here about the infinite sequence of solutions of the Diophantine equation $\displaystyle a^2+1 = 2b^2$. Those solutions obviously satisfy the equation $\displaystyle (7a)^2 + 49 = 2c^2$, where $\displaystyle c=7b.$ If you then put $\displaystyle x = \frac{7(a-1)}2 = \frac12\bigl(-7+\sqrt{2c^2-49}\bigr)$, you will have an infinite sequence of solutions of the equation $\displaystyle x^2 + (x+1)^2 = c^2.$

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. - Dec 16th 2010, 08:27 AMWilmerHmmm....
- Dec 17th 2010, 03:16 AMribbie
I'm not sure I understand this. If $\displaystyle x = \frac{7(a-1)}2 $ then I can use $\displaystyle x = 35$, for example, but that gives me http://www.mathhelpforum.com/math-he...603a864af1.png, where $\displaystyle c^2

= 2989$ but that means that c is not a whole integer (which is what I am trying to find.) - Dec 17th 2010, 06:41 AMOpalg
No, you can't use $\displaystyle x=35$ because that would come from $\displaystyle a=11$, which is not an allowable value of $\displaystyle a.$ In fact, $\displaystyle a$ must be one of the Pell numbers $\displaystyle H_{2n+1} = \frac12\bigl((\sqrt2+1)^{2n+1} - (\sqrt2-1)^{2n+1}\bigr).$ Those numbers don't look as though they will be integers, but when you expand the binomial series for $\displaystyle (\sqrt2\pm1)^{2n+1}$, the terms with $\displaystyle \sqrt2$ all cancel out, leaving you with an integer. The first few such integers are $\displaystyle a = 1,\ 7,\ 41,\ 239,\ 1393,\ 8119,\, ...\,.$ When you form $\displaystyle x = \frac72(a-1)$, you get the numbers $\displaystyle x = 0,\ 21,\ 140,\ 833,\ 4872,\ 28413,\, ...\,.$ In each case, you can check that $\displaystyle x^2+(x+7)^2$ is a square. As I said in my previous comment, you don't get anything like all the possible values of $\displaystyle x$ in that way. But I think that you do get all those that are multiples of 7.