Equation soluble (number theory)

[Funky1981]

Homework Statement

p is an odd prime

(a) show that x^2+y^2+1=0 (mod p) is soluble

(b) show that x^2+y^2+1=0 (mod p) is soluble for any squarefree odd m

Homework Equations

For (a) hint given : count the integers in {0,1,2,...,p-1} of the form x^2 modulo p and those of the form -1-y^2 modulo p

can anyone help me ?? thanks!

 

[Dick]

Homework Statement

p is an odd prime

(a) show that x^2+y^2+1=0 (mod p) is soluble

(b) show that x^2+y^2+1=0 (mod p) is soluble for any squarefree odd m

Homework Equations

For (a) hint given : count the integers in {0,1,2,...,p-1} of the form x^2 modulo p and those of the form -1-y^2 modulo p

can anyone help me ?? thanks!

Ok, here's another hint. If a^2 and b^2 are the same mod p, then a^2-b^2=0 mod p. So (a-b)(a+b)=0 mod p. What can you conclude about the relation between a and b and why?

 

[Funky1981]

Ok, here's another hint. If a^2 and b^2 are the same mod p, then a^2-b^2=0 mod p. So (a-b)(a+b)=0 mod p. What can you conclude about the relation between a and b and why?

thanks, i have solved the (a) part. For (b), I got the idea that since m is squarefree odd, so i write m = p1p2p3... (pi are prime) then by (a) the equation has solution for every pi. but how can i conclude that it has solution for their product?( btw , i made a mistake for typing, for (b) it should be modulo m but not p)

 

[Dick]

I see the OP got a hint on another forum. You can use the Chinese Remainder Theorem to take the solutions for p1, p2, ... and use them to construct a solution for the case mod m. Wish I had thought of it.