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Old 2020-09-09, 02:50   #1
bhelmes
 
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Mar 2016

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Default 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
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Old 2020-09-09, 06:30   #2
Batalov
 
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Mar 2008
Phi(4,2^7658614+1)/2

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All you need is ü(x,y) = x2-y2 (and throw away f() and g() ).
ü(x,y) "covers" all primes > 2
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Old 2020-09-09, 12:15   #3
Dr Sardonicus
 
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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).
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Old 2020-09-09, 18:31   #4
bhelmes
 
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Thanks for this clear answer.
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