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Equation soluble (number theory)

  • Thread starter Funky1981
  • Start date
  • #1
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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!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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?
 
  • #3
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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 eqation 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)
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
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