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sum of the squares of primes equalling a square

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  • Mark
    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
    Message 1 of 7 , Jul 6 8:26 AM
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      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)

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

      (For example 2^2 + 2^2 + 2^2 + 3^2 + 5^2 + 13^2 + 19^2 = 24^2 )

      Mark
    • 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 2 of 7 , Jul 6 8:53 AM
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        --- 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 3 of 7 , Jul 6 9:12 AM
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          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 4 of 7 , Jul 6 10:42 AM
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            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 5 of 7 , Jul 6 2:35 PM
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              --- 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 6 of 7 , Jul 6 3:34 PM
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                --- 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 7 of 7 , Jul 7 6:02 AM
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                  > 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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