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 oe (say) 900625... The series could be and probably is infinite anyway. --Pete Lomax (talk) 14:53, 21 December 2021 (UTC)