 By Greco S., Strano R.

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Extra resources for Complete Intersections

Example text

We say that the integers m and n are coprime (or relatively prime) if gcd(m,n) = 1. In this case, the associated illustration of the Euclidean algorithm finishes with the smallest square 1 x 1. Note that two consecutive integers are coprime. 4. Let d = gcd(m, n) for some integers m and n not both equal to zero. Then there exist integers x and y such that mx+ny = d. Proof. Let S be the set of all integers of the form ma + nb, where a and b are integers. Since one of m or n is not zero, there are nonzero integers in S.

R. Let ei = ordNai. Then ei I N -1 and ei (N -l)/Pi. Therefore, i I ei. 13, ei I ¢(N). Consequently, I ¢(N) for i = 1, ... , r. 4. 3 is proved for the case in which is fixed for 1 i T. Thus, ai is a primitive root modulo N in this instance. :s :s As promised earlier, we now present Pepin's interesting and useful test for the primality of the Fermat numbers. , in [Dickson], [Lenstra, Lenstra, Manasse, Pollard]. 5 (Pepin's Test). 1 ) Proof. 1) holds if m claim that the Jacobi symbol satisfies = 1.

24 17 lectures on Fermat numbers Proof. 26) is clearly satisfied. 27) 1 - 1 = (a(P-l)/2 - 1) (a(P-l)/2 + 1) := 0 f a. By (mod pl. Since p is a prime, either a(p-l)/2 := 1 (mod p) or a(p-l)/2 := -1 (mod pl. It thus suffices to prove that a(p-l)/2 := 1 (mod p) if and only if a is a quadratic residue modulo p. , a := b2 (mod p) for some integer b such that b -=f'. 0 (mod pl. Then by Fermat's little theorem, Conversely, assume that a(p-l)/2 := 1 (mod pl. Let q be a primitive root modulo p. Then a:= qt (mod p) for some t such that 1 ::; t ::; p - 1.