Thu, 12 Jan 2006 23:15:26 -0500 (EST) Edwin Clark <eclark@math.usf.edu> On Thu, 12 Jan 2006, Michael Reid wrote:
too many questions, not enough answers!
How about this one? Does every positive integer K appear in some square? That is, given K are there integers x,y and N such that N^2 = xKy? Yes, unless my early-morning-brain is muddled. In any base b, let x be ceiling(log_b (K)), and let d = ceil(fractional_part(K/2)) [in other words if K is odd, then d is a base b digit > b/2, else d < b/2]. Now, let N = 1 + floor(K/2)*b^{x+2} + d*b^{x+1} + 1. N^2 will be (K/2)^2 shifted 2x+4, K will appear at the {x+2} position, possibly followed by a single digit, then there will be a long string of zeros ending in 1. Integers may be expressed in any base and xKy is concatenation of x, K and y. --Edwin _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun