A tridigital square is a square that contains at most three Joe Crump (email) was so kind to explain the quadratic
distinct or unique squares like for instance with digits {2,6,9}
62969966996922666266229669292996
which is the square of 7935361806302386.
residues technique that turned out to be very effective in
discovering more tridigital squares. I let him do the talking...
Basically I iteratively test numbers of the form (10^A)x + y, where A is a chosen number of digits. I build a list of candidate y's by testing all the possible trailing A-digit combinations that satisfy the tridigital property, and who are quadratic residues.... To explain quadratic residues consider the following. All numbers could be represented as one of the following forms: 10x+0, 10x+1, 10x+2, 10x+3, 10x+4, 10x+5, 10x+6, 10x+7, 10x+8, 10x+9 If we square all these forms we find that no square can end with digits 2, 3, 7, or 8. If we do the same thing for 100x+c, c=0..99, we find only the following can be the last two digits of squares... 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96 Quadratic residues allow you to quickly identify non squares. For this problem, you could have also just iterated the numbers up to 10^A and collected those whose square satisfied the tridigital property, but quadratic residues come in handy in several places. Anyway, I build of list of y, then for each y only test those x's where the leading digits could satisfy the tridigital property. For instance, let's say we're looking for squares with only digits 2,3,9 and we are testing numbers of the form 1000x+177. It only takes a glimpse at the calculations with increasing x to see that we would be wasting a lot of tests on numbers that couldn't possibly satisfy the equation. I.e. x=100, we have 10035431329, x=101 we have 10236785329, etc. The target must at least start with 22*, and so the leading digits of 1000x+177 must not exceed an equal number of digits in sqrt(20/9) (1.490711985). So, long story short, I'm using quadratic residues and leading significant digits to shorten the range of numbers to test. I have an idea though to possibly find HUGE solutions although I need to research it to see if it could be fruitful. We can arbitrarily find large numbers whose square starts with at least half of whatever digits we want. For instance, take the sqrt(20/9) = 1.490711985 and iteratively compute the square of the first x digits rounded, i.e. (14+1)^2 = 225 (149)^2 = 22201 (1491)^2 = 2223081 (14907)^2 = 222218649 (149071)^2 = 22222163041 etc. We have squares that lead with all 2's. Now, all we need is to make the trailing digits satisfy the property we're after or just iterate a ton of these numbers until such a one falls into place. Also, keep in mind that numbers close to these numbers also have the same property so there is a good range to test. Furthermore, you'll find that a lot of the solutions known are relatively close to the sqrt approximations for the fraction representing repeating digits. A good example is: 8819171038 squared = 77777777797497997444 The square root of 7/9 is 0.881917103688.... and you see how close it is to the target number. I'm excited about finding some possibly HUGE solutions (i.e. 100+ digits) this way, but so far haven't done much with it. I did verify though of the first 100 digit numbers the best solution I could find is the one above from sqrt(7/9). It shows promise though as it required comparitively little calculation. Take care! Joe Crump [ September 9, 2000 ]
References
http://www.immortaltheory.com/NumberTheory/TriDigitalSquares.htm
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/overview.htm
https://erich-friedman.github.io/mathmagic/0999.html
http://www.worldofnumbers.com/threedigits.htm
[ TOP OF PAGE]