polynomalgebra, solution for the zero points ?

bhelmes · 2018-03-02, 20:07

A peaceful evening for all,

if i have the function with two variable f(u,v)=u²+v² with u,v elem. of N
i could easily fix one variable and
regard the corresponding function with one variable f(u,4)=u²+16
i can determine the u with 5 | f(u,4) with u1=3 and u2=7 for example

i could use both fixed functions, either with fixed u or with fixed v
this is a lattice in two dimension, vertical and horizontal.

Is there a mathematical way to dertermine all 5 | f(u,v)
if i know the two solutions for one polynom f(u,4)

Would be nice if someone can give me a hint.

I tried to visualize that problem in order to have a better understanding:
http://devalco.de/poly_xx+yy_demo.php?show=5

Feel free to jump in the complex plane

Bernhard

VBCurtis · 2018-03-02, 22:08

Make a list of squares mod 5. The 1 mod 5 squares must match up with a square 4 mod 5, while 0 mod 5 squares must match up with another 0 mod 5 square.

1 mod 5:
1,16,36,81,121,196,256.etc
u = 1,4,6,9,11,14,16,etc

4 mod 5:
4,9,49,64,144,169,289,etc
v = 2,3,7,8,12,13,17,etc

Do you see a pattern?

bhelmes · 2018-03-03, 00:03

i got the idea:

all calculation mod 5, u²+v² = 0 mod 5

u = 0, 1, 2, 3, 4
v 0: x, 0, 0, 0, 0
1: 0, 0, x, x, 0
2: 0, x, 0, 0, x
3: 0, x, 0, 0, x
4: 0, 0, x, x, 0

calculation

u : 0,1,2,3,4
u²: 0,1,4,4,1
v²: 0,4,1,1,4
v : 0,2,1,1,2
--------3,4,4,3

thanks for the help.

This will speedup the calculation of the prime sieving for the function f(u,v)=u²+v² because i could sieve in two dimension with every prime.

Greeting from the euclidean norm

Bernhard

2018-03-02, 20:07	#1
bhelmes Mar 2016 3·5·23 Posts	polynomalgebra, solution for the zero points ? A peaceful evening for all, if i have the function with two variable f(u,v)=u²+v² with u,v elem. of N i could easily fix one variable and regard the corresponding function with one variable f(u,4)=u²+16 i can determine the u with 5 \| f(u,4) with u1=3 and u2=7 for example i could use both fixed functions, either with fixed u or with fixed v this is a lattice in two dimension, vertical and horizontal. Is there a mathematical way to dertermine all 5 \| f(u,v) if i know the two solutions for one polynom f(u,4) Would be nice if someone can give me a hint. I tried to visualize that problem in order to have a better understanding: http://devalco.de/poly_xx+yy_demo.php?show=5 Feel free to jump in the complex plane Bernhard

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2018-03-02, 22:08	#2
VBCurtis "Curtis" Feb 2005 Riverside, CA 1001011111101₂ Posts	Make a list of squares mod 5. The 1 mod 5 squares must match up with a square 4 mod 5, while 0 mod 5 squares must match up with another 0 mod 5 square. 1 mod 5: 1,16,36,81,121,196,256.etc u = 1,4,6,9,11,14,16,etc 4 mod 5: 4,9,49,64,144,169,289,etc v = 2,3,7,8,12,13,17,etc Do you see a pattern?

2018-03-03, 00:03	#3
bhelmes Mar 2016 3·5·23 Posts	i got the idea: all calculation mod 5, u²+v² = 0 mod 5 u = 0, 1, 2, 3, 4 v 0: x, 0, 0, 0, 0 1: 0, 0, x, x, 0 2: 0, x, 0, 0, x 3: 0, x, 0, 0, x 4: 0, 0, x, x, 0 calculation u : 0,1,2,3,4 u²: 0,1,4,4,1 v²: 0,4,1,1,4 v : 0,2,1,1,2 --------3,4,4,3 thanks for the help. This will speedup the calculation of the prime sieving for the function f(u,v)=u²+v² because i could sieve in two dimension with every prime. Greeting from the euclidean norm Bernhard