Talk:Steady squares: Difference between revisions
Why not 1001 or 900625?
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::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. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 15:49, 21 December 2021 (UTC)
:::Pete you were right.My first thought was, there where only a finite set of steady squares.<br>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.<br>I am happy for not writing "proof" ;-) --[[User:Horst.h|Horst.h]] ([[User talk: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 g<sub>2</sub>g<sub>1</sub>g<sub>0</sub>.
Uncontentiously I assert that g<sub>0</sub> is 1, 5 or 6. Prepending n=0..9 to each and squaring these to obtain xg<sub>1</sub>g<sub>0</sub> and filtering for g<sub>1</sub>=n, I obtain 01, 25, 76 which are candidates for the next iteration. Note that using n=0 for g<sub>0</sub>=5 excludes 05 because 0 does not equal 2 and that it is only necessary to check g1 against n. Extending for g<sub>3</sub> makes the candidate list 001, 625, and 376. At g<sub>4</sub> interestingly 0625 is included because g<sub>4</sub> in 390625 passes g<sub>4</sub>=n for n=0. Note 00625 will fail at the next iteration, thus eliminating 900625.
Considering g<sub>0</sub>=1 and g<sub>n-1</sub>..g<sub>1</sub>=0
<pre>
101* 201* 1001*
101 201 1001
----- --- ----
101+ 201+ 1001+
10100 20200 1001000
----- ----- -------
10201= 20401= 1002001=
</pre>
Then g<sub>n</sub> is always 2n%10 which never equals n.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 14:15, 23 December 2021 (UTC)
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