66 I. R. <strong>Shafarevich</strong>Problems1. Prove that there are innitely many primes of the form 3s +2.2. The same for the primes of the form 4s +3.3. Prove that each two numbers 2 2n +1 and 2 2m +1 are relatively prime.Deduce once more the innity ofthenumber of primes. [Hint. Assuming that p isa common divisor of two such numbers, nd the remainders of division of 2 2m and2 2n by p.]4. Let f(x) be a polynomial with integer coecients. Prove that there existinnitely many distinct prime divisors of its values f(1), f(2), . . . . (If you do notsucceed immediately, solve the problem for the polynomials of the rst and of thesecond degree.)5. Denote by p n the n-th prime in the natural order. Prove that p n+1
Selected chapters from algebra 674, . . . , where 1 stands on odd places, and natural numbers stand on even places insuccession, is unbounded, but not unboundedly increasing, since one can nd thenumber 1 arbitrarily far in it.If a sequence a = a 1 a 2 ...a n ... of positive numbers is given, and s = Sa,then s n+1 >s n (since s n+1 = s n +a n+1 , a n+1 > 0), and, generally, s m >s n for m>n. Therefore, such a sequence will be unboundedly increasing if it is unbounded.For example, if all a i =1,thens n = n and the sequence s 1 s 2 ... is unbounded.But in other cases it may be bounded.An example can be visualised on Fig. 1,where we rst divide the segmentbetween0 and 1 in half and set a 1 = 1 , then divide2again the segmentbetween0and 1 in half2and set a 2 = 1, etc. In this way, a 4 n =12 n . The result of adding such numbersis represented on Fig. 1 and it is obviousthat the sums S n stay inside our initialsegment, i.e., S n < 1. Fig. 1It is easy to check the last assertion by calculation. If a n = 12 n ,then(Sa) n = 1 2 + 1 4 + + 12 n = 1 1+ 1 2 2 + + 12 n;1and by formula (12) of Chapter I(Sa) n = 1 212 n ; 1 112 ; 1 =1; 2 n so that (Sa) n < 1 for each n.We shall show now that in the case of the sequence 1, 1, 1, ... the rst case2 3appears: although the terms of the sequence decrease, they do not decrease fastenough, and their sums (i.e., S ;1 (n)) increase unboundedly.LEMMA 1. The sum S ;1 (n) is, for n suciently large, greater than an arbitrarygiven number.Let the number k be given. We assert that for some n (and so also for allgreater integers) S ;1 (n) >k. Take n such thatn ; 1=2 m for some m. Divide thesum 1 1S ;1 (n) =1+ +2 3 + 1 1+4 5 + 1 6 + 1 7 + 1 1++8 2 m;1 +1 + + 12 min parts as it is shown: in groups contained between two consecutive powers of two.Each parenthesis has the form 1 + + 12 k;1 +1 2 k , and the number of parenthesesis equal to m. In each parenthesis we replace each summand by the smallest oneentering that parenthesis, that is by the last one. Since the numberofsummandsinsuch a parenthesis is equal to 2 k ;2 k;1 =2 k;1 ,we obtain that the k-th parenthesis