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Re: sum of the squares of primes equalling a square

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  • elevensmooth
    ... 8^2 + 49^2 + 64^2 = 81^2 For any integers p, q, s, and t let d=(p^2+q^2+s^2+t^2)/2 c=(p^2+q^2-s^2-t^2)/2 a=ps+qt b=pt-qs then a^2 + b^2 + c^2 = d^2
    Message 1 of 7 , Jul 6 8:53 AM
      --- In primenumbers@yahoogroups.com, "Mark" <mark.underwood@...> wrote:
      >
      > Some tidbits that some may find interesting and perhaps fun to prove:
      >
      > The sum of the squares of 3 primes never equals a square.

      8^2 + 49^2 + 64^2 = 81^2

      For any integers p, q, s, and t let

      d=(p^2+q^2+s^2+t^2)/2
      c=(p^2+q^2-s^2-t^2)/2
      a=ps+qt
      b=pt-qs

      then a^2 + b^2 + c^2 = d^2

      Depending on the choices, c and d might be halves, and all the values will need be doubled to get integers.

      William Lipp
      Poohbah of OddPerfect.org
    • Jack Brennen
      Yes, the sum of squares of 3 composites can equal a square. But Mark stated that the sum of squares of 3 primes cannot; it s pretty easy to prove. (Choose the
      Message 2 of 7 , Jul 6 9:12 AM
        Yes, the sum of squares of 3 composites can equal a square.

        But Mark stated that the sum of squares of 3 primes cannot;
        it's pretty easy to prove. (Choose the right modulus for
        solving your Diophantine equation and it becomes clear.)

        On 7/6/2012 8:53 AM, elevensmooth wrote:
        >
        >
        > --- In primenumbers@yahoogroups.com, "Mark" <mark.underwood@...> wrote:
        >>
        >> Some tidbits that some may find interesting and perhaps fun to prove:
        >>
        >> The sum of the squares of 3 primes never equals a square.
        >
        > 8^2 + 49^2 + 64^2 = 81^2
        >
        > For any integers p, q, s, and t let
        >
        > d=(p^2+q^2+s^2+t^2)/2
        > c=(p^2+q^2-s^2-t^2)/2
        > a=ps+qt
        > b=pt-qs
        >
        > then a^2 + b^2 + c^2 = d^2
        >
        > Depending on the choices, c and d might be halves, and all the values will need be doubled to get integers.
        >
        > William Lipp
        > Poohbah of OddPerfect.org
        >
        >
        >
        >
        > ------------------------------------
        >
        > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
        > The Prime Pages : http://primes.utm.edu/
        >
        > Yahoo! Groups Links
        >
        >
        >
        >
        >
      • Mark
        True. And those identities from William were pretty neat. I had said, If the sum of the squares of 5 primes equals a square, two of the primes are 2 and 3
        Message 3 of 7 , Jul 6 10:42 AM
          True. And those identities from William were pretty neat.

          I had said,

          "If the sum of the squares of 5 primes equals a square, two of the primes are 2
          and 3"

          which was a little vague. It should really have been

          If the sum of the squares of 5 primes equals a square, one of the primes is 2 and another is 3.

          Besides the previous

          If the sum of the squares of 7 primes equals a square, three of the primes are
          2.

          we now have this:

          If the sum of the squares of 11 primes equals a square, two of the primes are 2 and another is 3.

          I never would have guessed off the top of my head that with this many squares there would be such restrictions!

          Mark



          --- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
          >
          > Yes, the sum of squares of 3 composites can equal a square.
          >
          > But Mark stated that the sum of squares of 3 primes cannot;
          > it's pretty easy to prove. (Choose the right modulus for
          > solving your Diophantine equation and it becomes clear.)
          >
          > On 7/6/2012 8:53 AM, elevensmooth wrote:
          > >
          > >
          > > --- In primenumbers@yahoogroups.com, "Mark" <mark.underwood@> wrote:
          > >>
          > >> Some tidbits that some may find interesting and perhaps fun to prove:
          > >>
          > >> The sum of the squares of 3 primes never equals a square.
          > >
          > > 8^2 + 49^2 + 64^2 = 81^2
          > >
          > > For any integers p, q, s, and t let
          > >
          > > d=(p^2+q^2+s^2+t^2)/2
          > > c=(p^2+q^2-s^2-t^2)/2
          > > a=ps+qt
          > > b=pt-qs
          > >
          > > then a^2 + b^2 + c^2 = d^2
          > >
          > > Depending on the choices, c and d might be halves, and all the values will need be doubled to get integers.
          > >
          > > William Lipp
          > > Poohbah of OddPerfect.org
          > >
          > >
          > >
          > >
          > > ------------------------------------
          > >
          > > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
          > > The Prime Pages : http://primes.utm.edu/
          > >
          > > Yahoo! Groups Links
          > >
          > >
          > >
          > >
          > >
          >
        • Phil Carmody
          ... Most likely attack is mod 24. Squares are {0,1,4,9,16,12}. Prime squares are {1} and the trivial (choice-free) {4,9} So we have some 2s, some 3s, and then
          Message 4 of 7 , Jul 6 2:35 PM
            --- On Fri, 7/6/12, Mark <mark.underwood@...> wrote:
            > Some tidbits that some may find interesting and perhaps fun to prove:
            >
            > The sum of the squares of 3 primes never equals a square.
            >
            > If the sum of the squares of 5 primes equals a square, two
            > of the primes are 2 and 3.
            >
            > (For example 2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 16^2)

            Most likely attack is mod 24.
            Squares are {0,1,4,9,16,12}.
            Prime squares are {1} and the trivial (choice-free) {4,9}

            So we have some 2s, some 3s, and then some things that square to 1:

            \ 0*2 1*2 2*2 3*2 4*2 5*2
            0*3 0+5 4+4 8+3 12+2 16+1 20+0 | no solns
            1*3 9+4 13+3 17+2 21+1 1+0 x | 13+3 and 1+0 are squares
            2*3 18+3 22+2 2+1 6+0 x x | 22+2 is a square
            3*3 3+2 7+1 11+0 x x x | no solns
            4*3 12+1 16+0 x x x x | 16+0 is a square
            5*3 21+0 x x x x x | no solns

            13+3 is {2, 3, p, q, r}
            1+0 is {2, 2, 2, 2, 3}
            22+2 is {2, 3, 3, p, q}
            16+0 is {2, 3, 3, 3, 3}

            Your result is confirmed.

            > If the sum of the squares of 7 primes equals a square, three
            > of the primes are 2.

            Presuming the above attack works too:


            \ 0*2 1*2 2*2 3*2 4*2 5*2 6*2 7*2
            0*3 0+7 4+6 8+5 12+4 16+3 20+2 0+2 4+2 | 12+4 has 3*2
            1*3 9+6 13+5 17+4 21+3 1+2 5+1 9+0 x | 21+3 and 9+0 have >=3*2
            2*3 18+5 22+4 2+3 6+2 10+1 14+0 x x | no solns
            3*3 3+4 7+3 11+2 15+1 19+0 x x x | 15+1 has 3*2
            4*3 12+3 16+2 20+1 0+0 x x x x | 0+0 has 3*2
            5*3 21+2 1+1 5+0 x x x x x | no solns
            6*3 6+1 10+0 x x x x x x | no solns
            7*3 15+0 x x x x x x x | no solns

            Confirmed.

            Whenever there's a puzzle involving squares, 24 is a very powerful lever.

            Phil
          • djbroadhurst
            ... Phil wisely chose to use 24 = 2^3 * 3 for a modular exhaustion. When I first encountered number theory, it took me a while to understand why working mod 8
            Message 5 of 7 , Jul 6 3:34 PM
              --- In primenumbers@yahoogroups.com,
              Phil Carmody <thefatphil@...> wrote:

              > Most likely attack is mod 24.
              ....
              > Your result is confirmed.

              Phil wisely chose to use 24 = 2^3 * 3
              for a modular exhaustion.

              When I first encountered number theory,
              it took me a while to understand why
              working mod 8 was important, when it seemed
              to me (wrongly) that we need only care
              about mod 2.

              If one will glance, for example, at
              http://en.wikipedia.org/wiki/Kronecker_symbol
              then one will see that mod 8 criteria really matter.
              [Also when trying to extract modular square roots.]

              Silly summary: when dealing with any even prime,
              it may be wise to work modulo its cube.

              David
            • Mark
              ... Thank you Phil, that was very helpful. Now if it could only level a house made from hypothetical perfect euler bricks! (That happens to be the problem
              Message 6 of 7 , Jul 7 6:02 AM
                > Whenever there's a puzzle involving squares, 24 is a very powerful lever.

                Thank you Phil, that was very helpful. Now if it could only level a house made from hypothetical perfect euler bricks! (That happens to be the problem that led me down this particular sum of squares of primes theme, albeit as a side diversion!)


                Mark


                --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
                >
                > --- On Fri, 7/6/12, Mark <mark.underwood@...> wrote:
                > > Some tidbits that some may find interesting and perhaps fun to prove:
                > >
                > > The sum of the squares of 3 primes never equals a square.
                > >
                > > If the sum of the squares of 5 primes equals a square, two
                > > of the primes are 2 and 3.
                > >
                > > (For example 2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 16^2)
                >
                > Most likely attack is mod 24.
                > Squares are {0,1,4,9,16,12}.
                > Prime squares are {1} and the trivial (choice-free) {4,9}
                >
                > So we have some 2s, some 3s, and then some things that square to 1:
                >
                > \ 0*2 1*2 2*2 3*2 4*2 5*2
                > 0*3 0+5 4+4 8+3 12+2 16+1 20+0 | no solns
                > 1*3 9+4 13+3 17+2 21+1 1+0 x | 13+3 and 1+0 are squares
                > 2*3 18+3 22+2 2+1 6+0 x x | 22+2 is a square
                > 3*3 3+2 7+1 11+0 x x x | no solns
                > 4*3 12+1 16+0 x x x x | 16+0 is a square
                > 5*3 21+0 x x x x x | no solns
                >
                > 13+3 is {2, 3, p, q, r}
                > 1+0 is {2, 2, 2, 2, 3}
                > 22+2 is {2, 3, 3, p, q}
                > 16+0 is {2, 3, 3, 3, 3}
                >
                > Your result is confirmed.
                >
                > > If the sum of the squares of 7 primes equals a square, three
                > > of the primes are 2.
                >
                > Presuming the above attack works too:
                >
                >
                > \ 0*2 1*2 2*2 3*2 4*2 5*2 6*2 7*2
                > 0*3 0+7 4+6 8+5 12+4 16+3 20+2 0+2 4+2 | 12+4 has 3*2
                > 1*3 9+6 13+5 17+4 21+3 1+2 5+1 9+0 x | 21+3 and 9+0 have >=3*2
                > 2*3 18+5 22+4 2+3 6+2 10+1 14+0 x x | no solns
                > 3*3 3+4 7+3 11+2 15+1 19+0 x x x | 15+1 has 3*2
                > 4*3 12+3 16+2 20+1 0+0 x x x x | 0+0 has 3*2
                > 5*3 21+2 1+1 5+0 x x x x x | no solns
                > 6*3 6+1 10+0 x x x x x x | no solns
                > 7*3 15+0 x x x x x x x | no solns
                >
                > Confirmed.
                >
                > Whenever there's a puzzle involving squares, 24 is a very powerful lever.
                >
                > Phil
                >
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