User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T16:04:45Z http://mathoverflow.net/feeds/user/29817 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115149/is-there-a-composite-number-that-satisfies-these-conditions/116843#116843 Answer by unknown (yahoo) for Is there a composite number that satisfies these conditions? unknown (yahoo) 2012-12-20T07:10:28Z 2012-12-20T07:10:28Z <p>if q=4k+1 and prime, then (a+bi)^q is (a+bi) and NOT (a-bi). the "-" is only for prime q=4k+3. your proof is wrong from the start, please recheck it. </p> http://mathoverflow.net/questions/115149/is-there-a-composite-number-that-satisfies-these-conditions/115954#115954 Answer by unknown (yahoo) for Is there a composite number that satisfies these conditions? unknown (yahoo) 2012-12-10T06:35:47Z 2012-12-10T07:06:10Z <p>As Mr. R. Gerbicz pointed out in the <a href="http://www.mersenneforum.org/showthread.php?t=17537" rel="nofollow">mersenne forum</a> an eventual counterexample for the base 3+2i must be 13-PRP (just multiply the equation with its conjugate). The first point to check is to make a list of pseudoprimes base 13 which are 3 (mod 4). I checked them to 10^10 and there is no counterexample which pass this test (a couple of them which are 1 (mod 4) passes the complex base test, but none of the 3 (mod 4)). However, the general opinion is that this test is a "hidden" multi-base PRP test, or a (n-1)(n+1) combined test, and as Mr. Tom Womack pointed out in that thread, if a couterexample exists, it must be HIGH (somewhere in 10^30 or so). </p>