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Find the solution of x^2≡3 mod 23(a) x≡±16 mod 23(b) x≡±13 mod 23(c) x≡±22 mod 23(d) x≡±7 mod 23The question was posed to me in exam.Asked question is from Overview in portion Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct choice is (a) x≡±16 mod 23

To EXPLAIN I would say: a=33^((23+1)/4)≡3^6≡1(QR and there is SOLUTION)

x ≡ ±3(23 + 1)/4 (mod 23) ≡±16i.e. x = 7 and 16



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