# looking for a formula to solve a particular pythagorean triple

• December 16th 2010, 02:14 AM
ribbie
looking 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, 02:51 AM
earboth
Quote:

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

1. Do you mean

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

2. If so solve the equation for x:

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

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.
• December 16th 2010, 03:10 AM
ribbie
Quote:

Originally Posted by earboth
1. Do you mean

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

Yes!

Quote:

I've found (by brute force) 1, 7, 13, 17, 35, 103
also 65 and 97... but I need a formula, please.

Quote:

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

I was thinking we could use the rule $a = n^2 - m^2, b = 2nm, c = n^2 + m^2$
but I got lost trying to figure it out.
• December 16th 2010, 07:30 AM
Opalg
There is a page of information here about the infinite sequence of solutions of the Diophantine equation $a^2+1 = 2b^2$. Those solutions obviously satisfy the equation $(7a)^2 + 49 = 2c^2$, where $c=7b.$ If you then put $x = \frac{7(a-1)}2 = \frac12\bigl(-7+\sqrt{2c^2-49}\bigr)$, you will have an infinite sequence of solutions of the equation $x^2 + (x+1)^2 = c^2.$

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.
• December 16th 2010, 09:27 AM
Wilmer
Hmmm....
Quote:

Originally Posted by Opalg
I think that in that way you will get all the solutions for which x is a multiple of 7.

I think you get 'em all, Opalg.
1st 5 are:
5,12,13*
8,15,17
21,28,35
48,55,73
65,72,97

*If c = 13: x = [-7 + SQRT(338 - 49)] / 2 = 5 ; mais oui?
• December 17th 2010, 04:16 AM
ribbie
Quote:

Originally Posted by Opalg
There is a page of information here about the infinite sequence of solutions of the Diophantine equation $a^2+1 = 2b^2$. Those solutions obviously satisfy the equation $(7a)^2 + 49 = 2c^2$, where $c=7b.$ If you then put $x = \frac{7(a-1)}2 = \frac12\bigl(-7+\sqrt{2c^2-49}\bigr)$, you will have an infinite sequence of solutions of the equation $x^2 + (x+1)^2 = c^2.$

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.

I'm not sure I understand this. If $x = \frac{7(a-1)}2$ then I can use $x = 35$, for example, but that gives me http://www.mathhelpforum.com/math-he...603a864af1.png, where $c^2
= 2989$
but that means that c is not a whole integer (which is what I am trying to find.)
• December 17th 2010, 07:41 AM
Opalg
Quote:

Originally Posted by ribbie
I'm not sure I understand this. If $x = \frac{7(a-1)}2$ then I can use $x = 35$, for example, but that gives me http://www.mathhelpforum.com/math-he...603a864af1.png, where $c^2
= 2989$
but that means that c is not a whole integer (which is what I am trying to find.)

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