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G n. Therefore Qn— 1 < n. But Qn— 1 is not divisible by any prime number < n (because these numbers are divisors of gM) and, as n ^ 4, the number gn—1 ^ g4— 1 = 5 > 1 has a prime divisor /? which must be ^ n. A fortiori, Qn—l}tn which gives a contradiction. Therefore Qn > n for n > 3. About the product Pn of all prime numbers ^ w we can prove that, for a natural number n, Pn < 4n (see Sierpinski [7], p. 396) and, for a natural number n Ξ> 29, P n > 2n. It has also been proved that, for a natural number n > 2, the sum of all prime numbers ^ n is > n.

E. the number p) has eleven digits. It is easy to prove that for integral « > 1 π(η—\) «—1 π(η) « where « is a prime number, and π(η— 1) π(ή) «—1 « where « is a composite number. We can prove in an elementary way that the ratio of π(«) to « tends to zero when « tends to infinity. η) = n for any natural n. It is easy to prove that there exist arbitrarily long sequences of natural numbers which contain no prime numbers. , and the last by m+l; thus they are all composite. For m = 100 the numbers would be gigantic, but between the prime numbers 370,261 and 370,373 there lie 111 successive composite numbers.

Thus in order that the natural number n > 1 be prime it is necessary and sufficient that the number (n—1)! + 1 be divisible by n. Theoretically, then, by only one division we can find out whether a number is prime or not. +1 has more than a hundred digits. +1 is divisible by/? 2 . For/? ^ 50,000, there are only three such numbers: 5, 13 and 563 (Fröberg [1]). We do not know if such numbers p are infinitely many. The theorems of Fermat and Wilson can be combined in the following theorem: If p is a prime number, then for every integer a the number a r(p—l)la is divisible by /?.