Discrete Math , Second Problem Set ( June 24 )
نویسنده
چکیده
Proof: For a contradiction, assume there are finitely many, p1, . . . , pn. Construct N = Πi=1pi. Then for any i, pi does not divide N + 1, so N + 1 is not divisible by any prime. This is a contradiction, since all numbers ≥ 2 are divisible by some prime. (This can be easily proved by induction.) Definition 1.4. We say that two numbers, a, and b are congruent mod m, or a ≡ b mod m if m | b− a. Lemma 1.5. If n ≡ −1 mod 4, then it must have a prime divisor ≡ −1 mod 4.
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تاریخ انتشار 2003