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But insead I counted prime numbers of the form 4N+1.
To my (sleepy) surprise, it seemed as if
Moreover, it seemed as if
By multiplying both sides of equations 1) through 4) by P^2M or Q^2M, M>0, we can obtain similar results for any integer power of P or Q.
Bh12 ( talk) 11:23, 2 June 2015 (UTC)
Thank you! The references you listed give proofs for P, P^2, Q, and Q^2 (with Q having no factor greater than squares of primes), showing they can be set to two squares that are relatively prime.
They also give proofs for higher powers of P and Q, but it is not clear whether two relatively prime squares can always be found - i.e., it is not clear whether (for a given P or Q for a power of 3 or higher) a solution can be found that is free of a common factor.
(I rechecked the case for 5^5 and found only two solutions (55*2+10^2, and 50^2+25^2), neither of which has relatively prime squares; i.e., there is a common factor (5^2 and 5^4) on both sides of the equation.)
So some of the original questions still stand. Bh12 ( talk) 22:27, 2 June 2015 (UTC)
Mathematics desk | ||
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< June 1 | << May | June | Jul >> | Current desk > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
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The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
But insead I counted prime numbers of the form 4N+1.
To my (sleepy) surprise, it seemed as if
Moreover, it seemed as if
By multiplying both sides of equations 1) through 4) by P^2M or Q^2M, M>0, we can obtain similar results for any integer power of P or Q.
Bh12 ( talk) 11:23, 2 June 2015 (UTC)
Thank you! The references you listed give proofs for P, P^2, Q, and Q^2 (with Q having no factor greater than squares of primes), showing they can be set to two squares that are relatively prime.
They also give proofs for higher powers of P and Q, but it is not clear whether two relatively prime squares can always be found - i.e., it is not clear whether (for a given P or Q for a power of 3 or higher) a solution can be found that is free of a common factor.
(I rechecked the case for 5^5 and found only two solutions (55*2+10^2, and 50^2+25^2), neither of which has relatively prime squares; i.e., there is a common factor (5^2 and 5^4) on both sides of the equation.)
So some of the original questions still stand. Bh12 ( talk) 22:27, 2 June 2015 (UTC)