A tridigital square is a square that contains at most three

distinct or unique squares like for instance with digits {2,6,9}

62969966996922666266229669292996

which is the square of 7935361806302386.

Joe Crump (email) was so kind to explain the *quadratic 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, whereAis a chosen number of digits. I build a list of candidatey's by testing all the possible trailingA-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+9If we square all these forms we find that no square can end with digits2, 3, 7, or 8. If we do the same thing for100x+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, 96Quadratic residues allow you to quickly identify non squares. For this problem, you could have also just iterated the numbers up to10^Aand collected those whose square satisfied the tridigital property, but quadratic residues come in handy in several places. Anyway, I build of list ofy, then for eachyonly test thosex's where the leading digits could satisfy the tridigital property. For instance, let's say we're looking for squares with only digits2,3,9and we are testing numbers of the form1000x+177. It only takes a glimpse at the calculations with increasingxto see that we would be wasting a lot of tests on numbers that couldn't possibly satisfy the equation. I.e.x=100, we have10035431329,x=101we have10236785329, etc. The target must at least start with22*, and so the leading digits of1000x+177must not exceed an equal number of digits insqrt(20/9) (1.490711985). So, long story short, I'm usingquadratic residuesandleading significant digitsto 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 thesqrt(20/9) = 1.490711985and iteratively compute the square of the firstxdigits rounded, i.e.(14+1)^2 = 225 (149)^2 = 22201 (1491)^2 = 2223081 (14907)^2 = 222218649 (149071)^2 = 22222163041etc. We have squares that lead with all2'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:8819171038squared= 77777777797497997444The square root of7/9is0.881917103688.... 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 fromsqrt(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

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Patrick De Geest - Belgium - Short Bio - Some Pictures

E-mail address : pdg@worldofnumbers.com