From: cet1@cus.cam.ac.uk (Chris Thompson) Newsgroups: sci.math Subject: Re: Bi-digit Squares (base 10) Date: 1 Apr 1995 23:39:58 GMT Keywords: Squares, composed of two unique digits In article <3lcl0p\$1i5s@coyote.csusm.edu>, randall@coyote.csusm.edu (Randall Rathbun) writes: |> Has anyone done any more research into bi-digit squares, or those composed |> of two unique digits? A computer search quickly located the following |> non-trivial ones... |> n n^2 |> 4 16 |> 5 25 [snip] |> 264 69696 |> 3114 9696996 |> 81619 6661661161 |> |> (n=100,200,300,1000,2000,3000,.. etc. are considered trivial) I would count n=10,20,30 as "trivial" as well. |> are there any others? If so, the square must have more than 15 digits if it |> does exist. Squares composed of three digits seem much more common. |> If anyone has found additional results, could you please post? We had a long thread involving squares whose digits were in {1,4,9} a few years ago. I will dig it out my archives if you are interested. The "natural heuristic hypothesis" if you are looking for squares whose representation in base b has digits in a set A, |A| = a, is that there will be finitely many if a < \sqrt{b} and infinitely many if a > \sqrt{b}, because there are a^n candidates less than b^n with a "probability" of being square around b^{-n/2}. One has to eliminate parametric solutions like your {1,4,9}.10^{2m} to make this work, of course. (There are less trivial parametric solutions for a=3,b=10: consider for example 38^2, 338^2, 3338^2, ...). For a=2, b=10 the solutions die out pretty rapidly, and it seems very likely that 81619^2 is the largest. The fun with a=3, b=10 is that although one expects the solutions to die out, they do so very slowly, as (9/10)^m goes only slowly to zero. Although some extremely large squares with A={1,4,9}, b=10 were found, I don't think we ever got to the point were the "expected" number of larger ones was < 1 ! AFAIK, there are absolutely no non-trivial theoretical results in this area. Computationally, one can improve on brute force search, of course. Suppose b and A are given, and we are looking for n-digit squares. One can work from the high-order end: after choosing n/2 digits, the square root of the potential square is completely determined, and we just square it and check whether the remaining digits are in A or not. Or one can work from the low-order end, working out the possible square roots mod b^{n/2} of various n/2-digit endings, squaring them, and checking the resulting top halves. Either of these methods takes time about a^{n/2}. By working from *both* ends, one can reduce this to a^{n/3}: I never did get round to programming this, although others did, and got the largest solutions that way. Chris Thompson Internet: cet1@cus.cam.ac.uk JANET: cet1@uk.ac.cam.cus