Check if an integer is a perfect square

The code is right here, see its length guy:

uses crt;

type bigNum = string; xau = array[1..1000000] of bigNum;

var t, so, temp: bigNum; pri: xau; kiemtra, i, dem, c, dai, dem1: longint;

    bool: boolean;

function cmp(a,b: bigNum): integer;

   begin

      while length(a)<length(b) do a:='0'+a;

      while length(b)<length(a) do b:='0'+b;

      if a>b then exit(1);

      if a<b then exit(-1);

      exit(0);

   end;

function add(a,b: bigNum): bigNum;

   var carry, i, sum: longint; c: string;

   begin

      carry:=0; c:='';

      while length(a)<length(b) do a:='0'+a;

      while length(b)<length(a) do b:='0'+b;

      for i:=length(a) downto 1 do

         begin

            sum:=ord(a[i])+ord(b[i])+carry-96;

            carry:=sum div 10;

            c:=chr(sum mod 10 + 48)+c;

         end;

         if carry>0 then c:='1'+c;

      add:=c;

   end;

function sub(a, b: bigNum): bigNum;

  var c: bigNum; s, borrow, i: longint;

  begin

    borrow:=0; c:='';

    while length(b)<length(a) do b:='0'+b;

    for i:=length(a) downto 1 do

       begin

          s:=ord(a[i])-ord(b[i])-borrow;

          if s<0 then

             begin

                s:=s+10;

                borrow:=1;

             end else borrow:=0;

             c:=chr(s+48)+c;

       end;

    while (length(c)>1) and (c[1]='0') do delete(c,1,1);

    sub:=c;

  end;

Function bigDiv2(a,b:bigNum):bigNum;

var c, hold:bigNum;

    kb:array[0..10] of bigNum;

    i,k:longint;

 begin

     kb[0]:='0';

     for i:=1 to 10 do

        kb[i]:=add(kb[i-1],b);

     hold:='';

     c:='';

     for i:=1 to length(a) do

       begin

          hold:=hold+a[i];

          k:=1;

          while cmp(hold,kb[k])<>-1 do

          inc(k);

          c:=chr(k-1+48);

          hold:=sub(hold,kb[k-1]);

       end;

     while (length(c)>1) and (c[1]='0') do delete(c,1,1);

     bigDiv2:=c;

 end;

function bigMod2(a, b: bigNum): bigNum;

  var hold: bigNum; kb: array[0..10] of bigNum; i, k: longint;

  begin

     kb[0]:='0';

     for i:=1 to 10 do kb[i]:=add(kb[i-1],b);

     hold:='';

     for i:=1 to length(a) do

        begin

           hold:=hold+a[i];

           k:=1;

           while cmp(hold, kb[k])<>-1 do

           inc(k);

           hold:=sub(hold, kb[k-1]);

        end;

     bigMod2:=hold;

  end;

procedure Eratosthene(n: bigNum; var prime: xau; dodai: longint);

var i, tmp, tmp1, k, e: longint; j: bigNum;

begin

    fillchar(prime,sizeof(prime),'0'); tmp1:=length(n) div 2 + 1;

    prime[1]:='2'; prime[2]:='3'; i:=2; j:='3'; dodai:=2;

    while (length(prime[i])<tmp1) and (i<=1000000) and (length(j)<tmp1) do

      begin

        j:=add(j,'1');

        tmp:=length(j) div 2 + 1; e:=0;

        for k:=1 to i do

           begin

              if (length(prime[k])<=tmp) and (bigMod2(j,prime[k])='0') then e:=1;

           end;

        if e=0 then begin inc(i); prime[i]:=j; inc(dodai); end;

      end;

end;

BEGIN

    CLRSCR;

    Writeln('   PROGRAM CHECKING IF AN INTEGER LESS THAN 10^100 ');

    Writeln('             IS A PERFECT SQUARE ');

    Write('   Enter the number needed to check: '); Readln(so);

    dem:=0;

    while (so[length(so)]='0') do begin inc(dem); delete(so,length(so),1); end;

    if odd(dem) then writeln('That number is not a perfect square') else

     if so='1' then writeln('That number is a perfect square') else

       begin

          delete(so, length(so)-dem, dem);

          kiemtra:=0; bool:=(so[length(so)]='2') or (so[length(so)]='3')

                or (so[length(so)]='7') or (so[length(so)]='8');

          if (so<>'1') and (kiemtra=0) then

             if bool then

                writeln('That number is not a perfect square') else

               begin

                 t:='1';

                 for c:=1 to (length(so) div 2 + 1) do t:=t+'0';

                 Eratosthene(t,pri,dai);

                 for c:=1 to dai do

                   begin

                     dem1:=0;

                     while bigMod2(so, pri[c])='0' do begin

                       so:=bigDiv2(so, pri[c]); inc(dem1);

                      end;

                     if odd(dem1) then begin kiemtra:=1; break;

                       end else kiemtra:=0;

                   end;

                 if kiemtra=1 then writeln('That number is not a perfect square')

                   else writeln('That number is a perfect square');

               end;

       end;

    readln

END.

---------- Post added at 09:03 PM ---------- Previous post was at 08:59 PM ----------

LuthfiHakim said:

If your code said 5 is perfect square, then your code definitely was wrong. Sqrt(5) returns 2.2360679775. I could not see why your code failed to detect the fraction part.

The number limitation is not only in Pascal, but also with any programming language. Since these kind of data type work in limited range only.

I know that we can take the square root of an integer, but only in case that this number is not greater than 10^20. However, I have to write the code for processing the numbers greater than 10^20 to, so the code has to be in common. I don't wanna divide in so many cases. It can take a lot of time.