250 Problems in Elementary Number Theory - Sierpinski (1970)

70 250 PROBLEMS IN NUMBER THEORYLet n be a given positive integer. We have obviously n!lJi(n!)-I, hencef{n!) = n! k + 1, where k is an integer. The absolute value of the polynomialhex) (which is of the order> 0) increases over all bounds with x; for sufficientlylarge n we shall have therefore If(n!)1 = In!k+ 11 > 1, and the numbern!k+l has a prime divisor p. In view of pln!k+l we must have p > n,and since pflt(n!), the congruence hex) = 0 (modp) is solvable for a primemodulus p > n. Since n is arbitrary, we deduce that the congruence fi(x)== 0 (modp), and also the congruence f(x) = 0 (modp) is solvable for infinitelymany primes p.131. There is only one such number, namely k = 1. Then the sequencek+l, k+2, ... , k+l0 (1)contains five primes: 2, 3, 5, 7, and 11. For k = 0 and k = 2,' sequence(1) contains four primes. If k ~ 3, then sequence (1) does not containnumber 3; as we know, out of each three consecutive odd numbers, one mustbe divisible by 3. It follows that sequence (1) contains at least one oddcomposite number. Besides that, sequence (1) contains five even numbers,hence (for k ~ 2) these numbers are composite. Thus, for k ~ 3, sequence(1) contains at least 6 composite numbers, and the numbers of primes cannotexceed 4.REMARK. Sequence (1) contains four primes for k = 0, 2, 10, 100, 190,820. We do not know whether there exist infinitely many such numbers k.From a certain conjecture of A. Schinzel concerning primes ([22]) it followsthat the answer is positive.132. There exists only one such number, namely k = 1. For this valuethe sequencek+l, k+2, ... , k+:lOO (1)contains 26 primes. For k = 0, 2, 3 or 4, sequence (1) contains 25 primes.Thus, we may assume that k ~ 5. Sequence (I) contains 50 even numbers,which for k > 1 are all composite. Next, it contains also SO successiveodd numbers, and since every three consecutive odd numbers contain onedivisible by 3, sequence (I) contains at least 16 numbers divisible by 3,which are all composite for k > 2.Let us compute now the number of terms of sequence (1) which are divisibleby 5, and neither by 3 or 2. All such numbers will be of the form