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Math Help - looking for a formula to solve a particular pythagorean triple

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    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.
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    Quote Originally Posted by ribbie View Post
    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.
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    Quote Originally Posted by earboth View Post
    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.
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    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.
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    Hmmm....

    Quote Originally Posted by Opalg View Post
    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?
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    Quote Originally Posted by Opalg View Post
    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.)
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  7. #7
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    Quote Originally Posted by ribbie View Post
    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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