 mersenneforum.org > Math f(x,y)=x²+y², g(x,y)=x²-2y², h(x,y)=?
 Register FAQ Search Today's Posts Mark Forums Read 2020-09-09, 02:50 #1 bhelmes   Mar 2016 3·53 Posts f(x,y)=x²+y², g(x,y)=x²-2y², h(x,y)=? A peaceful night for you, I am looking for the third biquadratic function, so that all primes are "covered" by these three functions. There is a relationship concerning the pyth. triples. I tried to illustrate this: f(x, y)=x²+y² http://www.devalco.de/poly_xx+yy_demo.php g(x,y)=x²-y²+2xy http://devalco.de/poly_xx+2xy-yy_demo.php What is the third biquadratic function ? (I would like to have h(x, 1)=2x²+1 on the diagonal line, but h(x,y)=x²+2y² does not look right : http://devalco.de/poly_xx+2yy_demo.php ) Sometimes I am a little bit blind, thanks in advance if you give me a hint.   Bernhard   2020-09-09, 06:30 #2 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×37×131 Posts All you need is ü(x,y) = x2-y2 (and throw away f() and g() ). ü(x,y) "covers" all primes > 2   2020-09-09, 12:15 #3 Dr Sardonicus   Feb 2017 Nowhere 22×32×149 Posts Note: x2-y2+2*x*y = (x - y)2 - 2*y2 If m and n are nonzero integers, and p is a prime that does not divide m*n, then at least of of m, n, and m*n is a quadratic residue (mod p). This is determined by p modulo |4*m*n|. In general, this does not translate neatly to quadratic forms, but with m = -1 and n = 2 we are fortunate. For odd primes p we have the following: p = x2 + y2 p == 1, 5 (mod 8) (-1 a quadratic residue) p = x2 + 2*y2 p == 1, 3 (mod 8) (-2 a quadratic residue) p = x2 - 2*y2 p == 1, 7 (mod 8) (+2 a quadratic residue) WRT x2 - 2*y2 we are even more fortunate because the "antimorph" 2*x2 - y2 represents the same primes. This is not always the case. For example, the form x2 - 3*y2 represents primes congruent to 1 (mod 12), while the "antimorph" 3*x2 - y2 represents primes congruent to 11 (mod 12).   2020-09-09, 18:31 #4 bhelmes   Mar 2016 5678 Posts Thanks for this clear answer.     Thread Tools Show Printable Version Email this Page

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