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Thread: Diophantine equation

  1. #1
    Forum Admin topsquark's Avatar
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    Diophantine equation

    I think Diophantine equations are in Number Theory...

    In any event. I am trying to solve 25 n^2 + (10 a - 264) n - (3a^2 - 4a + 52) - p^2 = 0 for n and p, where a is a constant positive integer.

    I wasn't able to figure out how to do it by hand and I couldn't find a way to get W|A to understand the "a".

    Any help is appreciated.

    -Dan
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    Re: Diophantine equation

    If we let m=25n+(5a-132),
    the equation can be written as

    (m-5p)(m+5p)=4 \left(4681-355 a+25 a^2\right)

    This says m-5p and m+5p are divisors of 4 \left(4681-355 a+25 a^2\right)
    Given a, we can solve for m and p and then find n

    This is not the best solution. Ideally we would like to find a formula for n and p in terms of a
    Last edited by Idea; Oct 29th 2017 at 04:23 AM.
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    Re: Diophantine equation

    One explicit solution

    n=-182+14a-a^2

    p=936-71 a+5 a^2
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    Forum Admin topsquark's Avatar
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    Re: Diophantine equation

    I had a thought last night after I posted this. As I was finishing off the typing I noticed that it didn't work if a is to be an integer. Here's the rundown of what I came up with. Maybe someone can fix the details.

    My strategy was to complete the square on n, make a redefinition to eliminate n from the LHS. This gives a Pell-like equation.

    What I did:
    25 n^2 + (10 a - 264) n - (3a^2 - 4a + 52) - p^2 = 0

    Completing the square gives:
    \left ( 5n - \frac{10a - 254}{10} \right ) ^2 - p^2 = f(a)

    where f(a) = 3 a^2 - 4a + 52 + \left ( \frac{10a - 264}{10} \right ) ^2

    Letting m = \frac{10 a - 264}{10} gives a condition on a: m must be an integer. But according to this m is not an integer for any a. So it doesn't work.

    -Dan
    Last edited by topsquark; Oct 29th 2017 at 10:27 AM.
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    Re: Diophantine equation

    Quote Originally Posted by topsquark View Post
    25 n^2 + (10 a - 264) n - (3a^2 - 4a + 52)
    Above has to be a square...(since it equals p^2).
    Keeping that in mind in a looper, I get numerous solutions; example:
    (a,n,p)
    1,15,42
    2,10,2
    5,-9,62
    6,10,18
    ....
    97,47,246

    58 solutions keeping a from 1 to 99
    and n from -99 to 99.

    Did I commit an unforgivable goof?
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  6. #6
    Forum Admin topsquark's Avatar
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    Re: Diophantine equation

    Quote Originally Posted by DenisB View Post
    Above has to be a square...(since it equals p^2).
    Keeping that in mind in a looper, I get numerous solutions; example:
    (a,n,p)
    1,15,42
    2,10,2
    5,-9,62
    6,10,18
    ....
    97,47,246

    58 solutions keeping a from 1 to 99
    and n from -99 to 99.

    Did I commit an unforgivable goof?
    Thanks. I was looking for something more analytical, though. Diophantine equations are far from my field of expertise. For all I know maybe it can't be done.

    -Dan
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    Re: Diophantine equation

    If we let m=25n+(5a-132),
    the diophantine equation can be rewritten as

    (m-5p)(m+5p)=4 \left(4681-355 a+25 a^2\right)

    Finding solutions involves finding divisors of the quantity

    u=4681-355a+25a^2 for a given a

    Look for pairs \left\{f_1,f_2\right\} such that f_1f_2=u, f_2\geq f_1 and f_1\equiv 4 mod 5

    Then let m=f_1+f_2

    example: a=1

    \left\{f_1,f_2\right\}=\left\{-4351,-1\right\} giving n=-169

    and

    \left\{f_1,f_2\right\}=\left\{19,229\right\} giving n=15
    Last edited by Idea; Oct 30th 2017 at 05:39 AM.
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    Forum Admin topsquark's Avatar
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    Re: Diophantine equation

    Oooooooooooh! That's clever. That answers my question nicely.

    Thanks to all!

    -Dan
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