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Equation with primes

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  • Adam Karasek
    Welcome I have writed with this problem recently, but I still can not solve it. I need to solve a equation: a^2+b^2=c^2 where a and c are primes and b is a
    Message 1 of 2 , Nov 1, 2003
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      Welcome

      I have writed with this problem recently, but I still can not solve it.
      I need to solve a equation:
      a^2+b^2=c^2
      where a and c are primes and b is a product of not more then four prime
      numbers. I have found that all solutions are:
      1) a=3 b=4=2*2 c=5
      2) a=5 b=12=2*2*3 c=13
      3) a=11 b=60=2*2*3*5 c=61
      I have also concluded that:
      a=2q+1
      b=2q^2+2q
      c=2q^2+2q+1
      and:
      b=(a^2-1)/2
      c=(a^2+1)/2
      But I still don't know how should I proof that these three solutions
      are the only one. How should I do proof it?

      I would be grateful for any help.
      Best Regards
      Adam Karasek



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    • Dan Dima
      ... still can not solve it. ... more then four prime ... three solutions ... Indeed these 3 solutions you found are all solutions for the equation. 1) a=3
      Message 2 of 2 , Nov 18, 2003
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        --- In primenumbers@yahoogroups.com, Adam Karasek
        <adam_karasek@y...> wrote:
        > Welcome
        >
        > I have writed with this problem recently, but I
        still can not solve it.
        > I need to solve a equation:
        > a^2+b^2=c^2
        > where a and c are primes and b is a product of not
        more then four prime
        > numbers. I have found that all solutions are:
        > 1) a=3 b=4=2*2 c=5
        > 2) a=5 b=12=2*2*3 c=13
        > 3) a=11 b=60=2*2*3*5 c=61
        > I have also concluded that:
        > a=2q+1
        > b=2q^2+2q
        > c=2q^2+2q+1
        > and:
        > b=(a^2-1)/2
        > c=(a^2+1)/2
        > But I still don't know how should I proof that these
        three solutions
        > are the only one. How should I do proof it?
        >
        > I would be grateful for any help.
        > Best Regards
        > Adam Karasek

        Indeed these 3 solutions you found are all solutions
        for the equation.
        1) a=3 b=4=2*2 c=5
        2) a=5 b=12=2*2*3 c=13
        3) a=11 b=60=2*2*3*5 c=61
        Your conclusions are also right.
        For any a < 13 prime there is no other solution.
        If a >= 13 prime:
        a=p , p-odd prime p>=13.
        b=(p^2-1)/2 , no more than 4 prime factors(might be
        equally few)
        c=(p^2+1)/2 , also odd prime
        Generally for p-odd prime:
        p mod10 = 1 or 3 or 7 or 9 , then
        p^2 mod10 = 1 or 9 .
        1). p^2 mod10 = 9 then
        c mod10 = 5 , c-odd prime , c > 5 and 5|c,
        contradiction
        no solution in this case.
        2). p^2 mod10 = 1 then
        b mod10 = 0
        so 10|b then 5|b
        b=2*((p-1)/2)*((p+1)/2),
        (p-1)/2,(p+1)/2 are 2 consecutive integers, then their
        product is divisible by 2, then (2*2)|b
        (p-1)/2,(p+1)/2, (p+3)/2 are 3 consecutive integers,
        then their product divisible by 3.
        (p+3)/2 is not divisible by 3, if it is, then p+3 is
        divisible by 6 p divisible by 3 also p prime
        contradiction
        ((p-1)/2)*((p+1)/2) divided by 3, then 3|b
        so (2*2*3*5)|b already 4 prime factors dividing b,
        60|b.
        but p>=13 then b=(p^2-1)/2 >= 84 > 60.
        there is at least one other prime factor dividing b
        and (b/60)
        so if p^2 mod10 = 1 , b is at least product of 5 prime
        factors
        does not correspond to initial conditions.
        no solution in this case also
        The problem is proved. (seems looking "known" to me)
        I hope this will help you.

        Best regards,
        Dan Dima


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