20200909, 02:50  #1 
Mar 2016
2^{2}·89 Posts 
f(x,y)=x²+y², g(x,y)=x²2y², 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)=x²y²+2xy http://devalco.de/poly_xx+2xyyy_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 
20200909, 06:30  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010101100110_{2} Posts 
All you need is ü(x,y) = x^{2}y^{2} (and throw away f() and g() ).
ü(x,y) "covers" all primes > 2 
20200909, 12:15  #3 
Feb 2017
Nowhere
2^{2}·7·179 Posts 
Note: x^{2}y^{2}+2*x*y = (x  y)^{2}  2*y^{2}
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 = x^{2} + y^{2} p == 1, 5 (mod 8) (1 a quadratic residue) p = x^{2} + 2*y^{2} p == 1, 3 (mod 8) (2 a quadratic residue) p = x^{2}  2*y^{2} p == 1, 7 (mod 8) (+2 a quadratic residue) WRT x^{2}  2*y^{2} we are even more fortunate because the "antimorph" 2*x^{2}  y^{2} represents the same primes. This is not always the case. For example, the form x^{2}  3*y^{2} represents primes congruent to 1 (mod 12), while the "antimorph" 3*x^{2}  y^{2} represents primes congruent to 11 (mod 12). 
20200909, 18:31  #4 
Mar 2016
2^{2}·89 Posts 
Thanks for this clear answer.
