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Noi Mathematics: solution of quadratic congruence equation

2022-06-24 17:30:00 dllglvzhenfeng

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5 In this case, we ask x^2≡a(mod p)(1) p Is an odd prime number k It's a positive integer. seek x^2≡p^2(mod p^k) How many solutions ?(2) a Is the complete square of an integer p Prime number seek x^2≡a(mod p) How many solutions ?(3) Prove that if x^2≡a(mod p) There are only two solutions that x^2≡a(mod p^k) Only _ Homework help

5 In this case, we ask x^2≡a(mod p)

(1) p Is an odd prime number k It's a positive integer. seek x^2≡p^2(mod p^k) How many solutions ?

(2) a Is the complete square of an integer p Prime number seek x^2≡a(mod p) How many solutions ?

(3) Prove that if x^2≡a(mod p) There are only two solutions that x^2≡a(mod p^k) There are only two solutions (k Is an integer )

(4) find 10 Odd primes p Satisfy p|x^2+5 (x Is an integer ) another : Can you make an inference about these prime numbers

(5) find 10 Odd primes p Satisfy p|x^2+1 (x Is an integer ) another : Can we infer from these prime numbers

(1) p Is an odd prime number k It's a positive integer. seek x^2≡p^2(mod p^k) How many solutions ?

k=1, There is a solution x≡0(mod p)

k=2, There is a solution x≡0(mod p)

k>2, Obviously positive and negative p There are two different solutions , If A Is a different solution , be (A^2,p^k)=(p^2,p^k)=p^2,

therefore A^2=z^2*p^2,(z,p)=1,A=z*p

be A^2-p^2=(z^2-1)p^2=mp^k,(m,p)=1,(z-1)(z+1)=m*p^(k-2),z-1 And z+1 There is only one possibility p Multiple

therefore z=n*p^(k-2)+/-1 m=n(np^(k-2)+/-2), obviously (n,p)=1

take n=0,1,2,3,...p-1,p+1,p+2,.2p-1,2p+1,2p+2,...3p-1.(p^2-1)*p -1

total 2+(p-1)*2*(p^2-1) A solution

(2) a Is the complete square of an integer p Prime number seek x^2≡a(mod p) How many solutions ?

(3) Prove that if x^2≡a(mod p) There are only two solutions that x^2≡a(mod p^k) There are only two solutions (k Is an integer )

(4) find 10 Odd primes p Satisfy p|x^2+5 (x Is an integer ) another : Can you make an inference about these prime numbers

(5) find 10 Odd primes p Satisfy p|x^2+1 (x Is an integer ) another : Can we infer from these prime numbers



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