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p,q : 2 distinct odd primes n=pq gcd(x,pq) =1

p,q : 2 distinct odd primes n=pq gcd(x,pq) =1. a. Show that x 1/2 (n) 1 (mod p) and x 1/2 (n) 1 (mod q) p = 2k +1 & q = 2l +1 gcd(x,p)=1 & gcd(x,q) =1 x 1/2 (n) = x 1/2(p-1)(q-1) = x 1/2 (2k)(2l) = x 2kl .

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p,q : 2 distinct odd primes n=pq gcd(x,pq) =1

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  1. p,q : 2 distinct odd primesn=pqgcd(x,pq) =1 a. Show that x1/2(n)1 (mod p) and x1/2(n)1 (mod q) p = 2k +1 & q = 2l +1 gcd(x,p)=1 & gcd(x,q) =1 x1/2(n) = x1/2(p-1)(q-1)= x1/2 (2k)(2l) = x2kl. x1/2(n)  x(2k)l  x(p-1)l  1l  1 (mod p)( since we have x(p-1) 1 (mod p) Fermat theory) x1/2(n)  x(2l)k  x(q-1)k  1k  1 (mod q) (x(q-1) 1 (mod q) )

  2. b. Show that x1/2(n)1 (mod n) x1/2(n)1 (mod p) x1/2(n) =ap +1 x1/2(n)1 (mod q)  ap+1  1 (mod q)  ap  0 (mod q)  a  0 (mod q ) since gcd(q,p) =1  a = bq Therefore , x1/2(n) = bpq + 1 = bn +1 or x1/2(n)  1 (mod n) C . If ed  1(mod 1/2(n) ) then xed  x (mod n) ed = c(1/2 (n)) + 1 xed  x c(1/2 (n)) + 1  x (1/2 (n))cx  1c x  x (mod n )

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