Talk:Steady squares

Reduce search range by observation of the results

By taking a look at the results,I tried to search for results by prepending a digit to n.
And there comes the solution:By prepending a digit to n, than n*n must end in n to be a steady square.
So only solutions of one digit before can be solutions. <lang pascal> function CalcSquare(n:LongInt;Pot10:byte); //pot10 is 10^(count of digits of n -1) //prepend one digit to n var

 dgt: LongInt;

begin

   //ex.: n= 5,Pot10 = 1, dgt= 2;-> n2= (2*10*1+5)^2 = 625
 For dgt := 1 to 9 do
 begin
   // n= 5 , dgt= 2*10 -> n2= 625
   n2 = sqr(dgt*10*Pot10+n);   
   n2 = dgt*dgt*100*Pot10*Pot10+2*dgt*10*pot10*n+n*n;

// term: dgt*dgt*100*Pot10*Pot10+2*dgt*10*pot10*n// always has ends in Pot10+1 "0" digits // so n*n must end in n, to be a steady square

 end;

</lang>

Note also that it must end in 1,5 or 6 (not 0 as any number ending in 0 squared ends with twice as many zeros).--Nigel Galloway (talk) 12:46, 21 December 2021 (UTC)

Ah, you beat me to it (I posted my Phix entry before reading this). I was hopeful, as you (Horst) seem to suggest, that we only have to look at the k-1 digit solutions on each iteration, and that way maybe even find all possible such numbers there could ever be, that is if any iteration found none we'd be done, but alas zero-fill totally spanners that illusion (eg 90625 off the back of 625). Oh, unless I've missed a logical reason why it is not worth checking (eg) 1001 or (say) 900625... Then again the series could be and probably is infinite anyway. --Pete Lomax (talk) 14:53, 21 December 2021 (UTC)

Nevermind that last bit:

1000557423423230896109004106619977392256259918212890625^2 = 10011151575673345586...92256259918212890625 (109 digits)

is a triple-zero-fill, and I've found examples all the way up to (a quite random) 548 digit n with a 1096 digit square. --Pete Lomax (talk) 15:49, 21 December 2021 (UTC)

Pete you were right.My first thought was, there where only a finite set of steady squares.
Thanks for showing, that prepending 1..x zeros ( and increasing Pot10 the same way ) to n, which is generating a stead square, lets n*n still be a steady square.
I am happy for not writing "proof" ;-) --Horst.h (talk) 10:32, 22 December 2021 (UTC)

Why not 1001 or 900625

Consideration of this question leads to an improvement of the algorithm. Consider a number of the form g₂g₁g₀. Uncontentiously I assert that g₀ is 1, 5 or 6. Prepending n=0..9 to each and squaring these to obtain xg₁g₀ and filtering for g₁=n, I obtain 01, 25, 76 which are candidates for the next iteration. Note that using n=0 for g₀=5 excludes 05 because 0 does not equal 2 and that it is only necessary to check g1 against n. Extending for g₃ makes the candidate list 001, 625, and 376. At g₄ interestingly 0625 is included because g₄ in 390625 passes g₄=n for n=0. Note 00625 will fail at the next iteration, thus eliminating 900625.

Considering g₀=1 and g_n-1..g₁=0

  101*    201*     1001*
  101     201      1001
-----     ---      ----
  101+    201+     1001+
10100   20200   1001000
-----   -----   -------
10201=  20401=  1002001=

Then g_n is always 2n%10 which never equals n.--Nigel Galloway (talk) 14:15, 23 December 2021 (UTC)

Excellent stuff. Likewise on the latter point we cannot "zero-fill" the single digit g₀=5 or g₀=6 because neither g₁ nor g_n+1 can ever be 0 (with either being sufficient grounds for permanent elimination). Going back to the first paragraph, another way to say that (as a guide rather than a proof) might be, using the triple-zero-fill example above:

    55742..90625^2 = ...6100055742..90625
100055742..90625^2 = ...1100055742..90625

The three zeros in the first line make it a zero-fill candidate for three rounds, after which the second line will either "take over the mantle" in the list of potential candidates, or there won't be anything that does. Or it might be more accurate to say it only becomes a candidate in three rounds time?

From my experiments so far I suspect the number of candidates at each iteration settles into a simple pattern, at least it does not seem to grow much, and in fact it may just be two candidates occasionally hopping over each other. --Pete Lomax (talk) 01:38, 24 December 2021 (UTC)

Oh, ffs, off course, how did I not spot that? It is just a 5-chain and a 6-chain, with the 1-chain dying instantly. There will only ever be and always will be either one or two n-digit steady square numbers ending in 5 or 6 or both, maybe with the odd skip when both leap-frog. DOH. --Pete Lomax (talk) 02:05, 24 December 2021 (UTC)