250 Problems in Elementary Number Theory - Sierpinski (1970)

114 250 PROBLEMS IN NUMBER THEORYwhich implies thatWe have, however,Xl IIl-Yl > ~-l~ 3n-1XI-YI(since we cannot have Xl = 2, for then we would have Yl = 1, and2"-112 11 +1, which is impossible). We would therefore have 3 n - 1 < 2 n +l,which is impossible for n ~ 3.237. We shall use the well-known formularZ+22+ ... +n2 = n(n+l)6(2n+l) .We have to find the least integer n > 1 such that n(n+l)(2n+l) = 6m 2where m is a positive integer. We shall distinguish six cases:1. n = 6k, where k is a positive integer. Our equation takes on theformk(6k+l) (12k+l) = m 2 •The factors on the left-hand side are pairwise relatively prime, hence theyall must be squares. If k = 1, then 6k + 1 is not a square. The next squareafter 1 is 4. If k = 4, we have 6k+ 1 = 52, 12k+ 1 = 7 2 , and consequently,for n ~ 6k = 24 the sum 12+22+ ... +242 is a square of a positiveinteger, 70.2. n = 6k+ 1, where k is a positive integer. In this case we have(6k+l) (3k+l) (2k+l) = m 2 ,and each of the numbers 2k+l, 3k+l, and 6k+l (which are pairwiserelatively prime) must be a square. The least k for which the number2k+l is a square is k = 4; in this case, however, we have n = 6k+l > 24.3. n = 6k+2, where k is an integer ~ o. We have in this case(3k+l) (2k+l) (12k+5) = m 2 ,and the numbers 3k+l, 2k+l, and 12k+5 (as pairwise relatively prime)must be squares. If we had k = 0, the number 12k+5 would not be a