250 Problems in Elementary Number Theory - Sierpinski (1970)

SOLUTIONS 39hence a+b+c ~ 5+7+9 ~ 21 > 17, which shows that 17 does not havethe desired property.49*. We shall present the proof based on an idea of A. Schinzel (see [19]).Let k denote a given positive integer and let m be the positive integer whoseexpansion into prime powers is m = qflq~2 ... q~'. Let I(x) = x(x+2k) andlet i denote one of the numbers 1, 2, ... , s. We cannot have qilx(x+2k) forall integer x since then for x = 1 we would have qil2k+ 1, and for x = -1we would have qd2k-l, and q;!(2k+l)-(2k-l) = 2, which is impossiblein view of qjl2k+ 1 (and, in consequence, qiI1). Therefore there exists aninteger Xj such that qi,r Xi(Xi+2k) = I(xj). By the Chinese remaindertheorem, there exists a positive integer Xo such that Xo == Xi (mod qj) fori = 1,2, ... , s, which yields I(xo) == I(xt) ~ 0 (mod qi) for i = 1,2, ... , s.We have therefore (f(xo), q;) = 1 for i ~ 1,2, ... , s, which (in view of theexpansion of m into prime factors) gives (f(xo), m) = 1, or (xo(xo+2k), m)= 1. Thus, if we put a = xo+2k, b = xo, we shall have 2k = a-b, where(a, m) = 1, (b, m) = 1, which proves the theorem.Since adding arbitrary multiples of m to a and b does notREMARK.change the fact that 2k = a-b and (ab, m) = 1, we proved that, for everym, every even number can be represented in infinitely many ways as a differenceof positive integers relatively prime with m.We do not know whether every even number is a difference of two primes.From a certain conjecture on prime numbers of A. Schinzel ([22]), it followsthat every even number can be represented as a difference of two l'rimesin infinitely many ways.'50*. We shall present the proof given by A. Rotkiewicz. If u" is the 'nthterm of the Fibonacci sequence. and if m and nare positive integers,then(um• u,,) = Um,,, (see [27, p. 280, problem 5]). Since Ul = 1, we see that if Pkdenotes the kth successive prime. then every two terms of the increasinginfinite sequenceare relatively prime. Instead of PI< we could take here 22k + 1 since it is wellknown that (2 2m + 1, 2 2 "+1) = 1 for positive integers m and n =I: m.51*. We know that every divisor> 1 of the number F" = 2 2 "+1(n = 1,2, ... ) is of the form 2,,+2k+ 1 where k is positive integer (see,. forinstance, [37, p. 343, Theorem 5]). Since for positive integers nand k we