250 Problems in Elementary Number Theory - Sierpinski (1970)

SOLUTIONS 87For s = 3, the equation has only one solution in increasing positive integerssince we must have Xl > 1, hence Xl ;::: 2, and if we had Xl ;::: 3, wewould have X2 ;::: 4, X3 ;::: 5, which is impossible since11111 1-+-+- ~ -+-+- < 1.Xl X2 X3 3 4 5We have, therefore, X2 = 3, hence X3 = 6, and consequently 13 = 1. On theother hand, h > 1 since the equation1 1 1 1-+-+-+-= 1Xl X2 X3 X4has in positive integers the solutions 2, 3, 7, 42 and 2, 3, 8, 24 (and alsoother solutions).We can, therefore, assume that s;::: 4. In this case the equation111-+-+ ... +-=1Xl X2 Xs-lhas at least one solution in increasing positive integers Xl < X2 < ... < Xs-loand then the numbers 11 = 2, 12 = 3, t3 = 6Xlo t4 = 6X2, ••• , tsH = 6X S-lwill be increasing positive integers, and will satisfy equation (1). This solutionwill be different than each of the Is solutions obtained previously sincethere all numbers were even, while here the number 3 is odd. Thus, we haveIsH;::: 1.+ 1, hence IsH > I. for s ;::: 3, which was to be proved.163. Let In = n(n+ 1)/2 denote the nth triangular number. We easilycheck that1 III-+-+-+-= 1.t2 t2 13 t3Thus, it suffices to assume that s is an integer;::: 5. If s is odd, that is,s = 2k-1 where k is an integer;::: 3, then we have1 1 1 k+l 2 2 2 2-+-+ ... +-+- = 2 3 +- 3 4 + ... + (k l)k +-k12 t3 tk- 1 tk •• -