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Is this question acceptable?: [Closed] Necessary congruences for c to be a prime number if a^2=b^2+c^2

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    Given a Pythagorean triplet a^2=b^2+c^2, if (c^2-1)/8=x, where x a non-zero natural number, what are the necessary numerical congruences for a=m^2+n^2 and c=m^2-n^2 so that c can be any prime number?
    • CommentAuthorgrp
    • CommentTimeNov 13th 2011
     

    This question might be appropriate for math.stackexchange, except it seems clear that you have not thought it through, or at least worded it carefully. Here is a version you might try on stackexchange:

    I am trying to find (not necessarily primitive) Pythagorean triples (a,b,c) such that a^2 = b^2 + c^2 and with c prime and odd. Using the parameterization where a=m^2 +n^2 and c =m^2 -n^2, are there necessary congruences needed for m and n?

    If you try this, you will probably be reprimanded for not trying c=(m+n)(m-n) and deducing the obvious. Even without using the parameterization involving m and n, trying (a+b)(a-b)=p*p for some prime p is elementary number theory that is not quite in the scope of MathOverflow. I strongly recommend not making any more posts to MathOverflow or to meta.mathoverflow until you have made a number of successfully received posts on math.stackexchange.

    Gerhard "Ask Me About System Design" Paseman, 2011.11.13

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    This question is not acceptable. Also, I am unable to understand why you said anything about x. Couldn't you demand that c be odd with absolute value greater than 1 without using an extra letter?

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    This simple equation, (c^2-1)/8=x, as far as I know, is the most effective in the whole of mathematics, for screening primes. From your responses it seems to me that you didn't put enough thought into it before criticizing. I would be delighted to see someone present the necessary congruences.
    • CommentAuthorquid
    • CommentTimeNov 14th 2011 edited
     

    If a number is odd it is 1, 3, 5, or 7 modulo 8. So its square is 1, 9 , 25, or 49 modulo 8. Since all of 9 and 25 and 49 are congruent 1 modulo 8, the square is in fact always 1 modulo 8. By contrast, the square of an even number is never congruent 1 modulo 8. So all the condition that (c^2-1) / 8 is non-zero integer tells you is that c is odd and c^2 is not 1.

    In other words: what Scott said.

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    The only reason I put the equation (c^2-1)/8 in my question was to formulate the sequence of all numbers with last digits 1,3,7,9 (the number 5 appears only once). This sequence contains primes and composite numbers. If we stop the sequence at 10^2 then all composite numbers up to that number are divisible by 3,7. If we stop the sequence at 10^3 then all composite numbers are divisible by 3,7 11,13,17,19,23,29,31 and for 10^6 we have 167 such divisors.The congruences I am looking for will reduce the divisions by two thirds. If we choose a number in the sequence up to 10^6, instead of doing 167 divisions we do only 56. The congruences you gave me are good only to formulate the sequence with numbers with last digits1,3,7,9, which is something I already knew.
    • CommentAuthorMariano
    • CommentTimeNov 14th 2011
     

    This thread should probably be closed...

    • CommentAuthorquid
    • CommentTimeNov 15th 2011
     

    @V.P.: Either you wrote something different from what you mean, or I do not know. In any case, I agree that it will be best if you'd stop asking on MO on this subject.

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    I'm closing this thread, as we seem to have come to the conclusion that this in not an appropriate question for MathOverflow.

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