Wikipedia:Reference desk/Archives/Mathematics/2015 June 2

= June 2 =

I Should Have Counted Sheep...
But insead I counted prime numbers of the form 4N+1.

To my (sleepy) surprise, it seemed as if
 * 1) P  =A^2+B^2
 * 2) P^2=C^2+D^2
 * where
 * P is a prime of form 4N+1;
 * P, A, and B are relatively prime integers;
 * P, C, and D are relatively prime integers.
 * Assuming this is not new, what is the proof - or is there a counterexample? A link will do.

Moreover, it seemed as if
 * 3) Q  =E^2+F^2
 * 4) Q^2=G^2+H^2
 * where
 * Q is the product of a number of primes of form 4N+1;
 * Q, E, and F are relatively prime integers;
 * Q, G, and H are relatively prime integers.
 * Again, assuming this is not new, what is the proof - or is there a counterexample? A link will do.

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.
 * However, these results have a common factor on both sides of the equations: P^2M or Q^2M.
 * So, finally, when is it possible to obtain results for integer powers >2 for P or Q but without a common factor on both sides?
 * For example, for P=5 or 13, or Q=65:
 * 5^3=11^2+2^2;
 * 13^3=46^2+9^2;
 * 65^3=524^2+7^2;
 * 5^4=24^2+7^2;
 * 5^6=117^2+44^2.

Bh12 (talk) 11:23, 2 June 2015 (UTC)


 * See Fermat's theorem on sums of two squares and Brahmagupta–Fibonacci identity. Gandalf61 (talk) 12:11, 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)