A course in number theory and cryptography
A course in number theory and cryptography
A course in computational algebraic number theory
A course in computational algebraic number theory
Finite fields
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Let p be an odd prime number and a a square modulo p. It is well known that the simple formula a \frac{p+1}{4} mod p gives a square root of a when p ≡ 3 mod 4. Let us write p − 1 = 2n s with s odd. A fast algorithm due to Shanks, with n steps, allows us to compute a square root of a modulo p. It will be shown that there exists a polynomial of at most 2n−1 terms giving a square root of a. Moreover, if there exists a polynomial in a representing a square root of a modulo p, it will be proved that this polynomial would have at least 2n−1 terms, except for a finite set \mathcal{P}n of primes p depending on n.