looking for a formula to solve a particular pythagorean triple

**ribbie** · December 16th 2010, 01:14 AM

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

**earboth** · December 16th 2010, 01:51 AM

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.

**ribbie** · December 16th 2010, 02:10 AM

Originally Posted by earboth

1. Do you mean

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

Yes!

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

also 65 and 97... but I need a formula, please.

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.

**Opalg** · December 16th 2010, 06:30 AM

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.

**Wilmer** · December 16th 2010, 08:27 AM

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?

**ribbie** · December 17th 2010, 03:16 AM

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

, where $c^2<br /> = 2989$ but that means that c is not a whole integer (which is what I am trying to find.)

**Opalg** · December 17th 2010, 06:41 AM

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

, where $c^2<br /> = 2989$ 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.

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