Loading ...
Sorry, an error occurred while loading the content.
 

y**2 + 5 x y + 5 x**2

Expand Messages
  • Kermit Rose
    ... Thank you. Now it is clear that 5 is indeed a quadratic residue of p = (137^137 + 1992)*(137^137 + 3464) And therefore, once we find the sqrt(5) mod p. we
    Message 1 of 2 , Oct 4 8:46 AM
      > ________________________________________________________________________
      > 1c. Re: qfbsolve exercise
      > Posted by: "djbroadhurst" d.broadhurst@... djbroadhurst
      > Date: Sun Oct 3, 2010 4:24 pm ((PDT))

      > --- In primenumbers@yahoogroups.com,
      > Kermit Rose<kermit@...> wrote:
      >
      >>> Exercise: Find two pairs of positive integers (x,y) such that
      >>> 5*x^2 + 5*x*y + y^2 = (137^137 + 1992)*(137^137 + 3464)
      >
      >> (137^137 + 1992)*(137^137 + 3464) = 1 * 3 = 3 mod 5.
      >

      > 137 = 2 mod 5
      > 137 = 1 mod 4
      > 137^137 = 2^1 mod 5
      > 137^137 + 1992 = 2 + 2 mod 5 = 4 mod 5
      > 137^137 + 3464 = 2 + 4 mod 5 = 1 mod 5
      >
      > David


      Thank you.

      Now it is clear that 5 is indeed a quadratic residue of
      p = (137^137 + 1992)*(137^137 + 3464)

      And therefore, once we find the sqrt(5) mod p.

      we need only find the x such that

      if y = [(5 + sqrt(5))/2 ] x mod p

      or y = [(5 - sqrt(5))/2 ] x mod p,


      then

      ( y + x [(5 + sqrt(5) )/2 ] ) ( y + x [ (5 - sqrt(5))/2 ] )

      = y**2 + 5 x y + 5 x**2

      = (137**137 + 1992)(137**137 + 3464)


      I presume that you constructed this puzzle by finding
      the first two primes > 137**137 which were prime and
      equal to either 1 or 4 mod 5.


      The smallest product of two distinct primes, = 1 or 4 mod 5,
      is 11 * 31 = 341

      4**2 = 5 mod 11
      6**2 = 5 mod 31

      t = 4 mod 11
      t = 6 mod 31

      37 = 4 mod 11
      37 = 6 mod 31

      37**2 = 1369
      1369 = 5 mod 341

      37**2 = 5 mod 341

      (5 + sqrt(5))/2 mod 341 = (5 + 37)/2 mod 341 = 21 mod 341

      (5 - sqrt(5))/2 MOD 341 = (5 - 37)/2 MOD 341 = 16 MOD 341


      Y = 21 X
      OR
      Y = 16 X

      (21 X)**2 + 5 X (21 X) + 5 X**2 = 341

      21**2 > 341

      (16 X)**2 + 5 x (16x) + 5 x**2 = 341

      256 x**2 + 80 x**2 + 5 x**2 = 341

      (256 + 80 + 5) x**2 = 341


      341 x**2 = 341

      x = 1

      y = 16

      16**2 + 5 * 1 * 16 + 5 * 1**2
      = 16**2 + 5 * 16 + 5
      = 256 + 80 + 5
      = 341.

      I expect that 341 is the smallest non-square composite positive integer, M,
      such that

      there exist x and y
      such that

      y**2 + 5 x y + 5 x**2 = M.



      Thank you Maximilian, for your comment,
      which implied that

      (137**137 + 1992)*(137**137 + 3464)

      is the product of two primes.

      >
      > there is no other factoring ;)
      >
      > Maximilian


      Kermit
    • Aldrich
      ... What about 11*19 = 209? quibble,quibble. a.
      Message 2 of 2 , Oct 5 10:11 AM
        --- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:
        >
        > > ________________________________________________________________________
        > > 1c. Re: qfbsolve exercise
        > > Posted by: "djbroadhurst" d.broadhurst@... djbroadhurst
        > > Date: Sun Oct 3, 2010 4:24 pm ((PDT))
        >
        > > --- In primenumbers@yahoogroups.com,
        > > Kermit Rose<kermit@> wrote:
        > >
        > > I expect that 341 is the smallest non-square composite positive integer, M,
        > such that
        >
        > there exist x and y
        > such that
        >
        > y**2 + 5 x y + 5 x**2 = M.
        >

        What about 11*19 = 209? quibble,quibble.

        a.
      Your message has been successfully submitted and would be delivered to recipients shortly.