Re: [math-fun] Digital silliness

On Fri, 13 Jan 2006, Michael B Greenwald wrote:

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}

That's nice! I figured there must be a simple way to do it. And if b is even, it is even simpler since then we can replace K by Kb and and let d = 0 so N is simply (K/2)*b^{x+2} + 1. At any rate, I think this indicates that solving David's problem may not be easy. --Edwin