Thread: looking for a formula to solve a particular pythagorean triple

1. 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.

2. 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.

3. 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.

4. 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.

5. Hmmm....

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?

6. 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 $c^2
= 2989$
but that means that c is not a whole integer (which is what I am trying to find.)

7. 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 $c^2
= 2989$
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.