Is it possible to form two numbers A and B from 22 digits such that A = B^2?

arithmetic/digits/squares/length.22 Added material

Problem

Is it possible to form two numbers A and B from 22 digits such that
A = B^2?  Of course, leading digits must be non-zero.

Show solution

No, the number of digits of A^2 must be of the form 3n or 3n-1.

Additional material

I find the following archive solution confusing, probably even
incorrect (although I'm not sure what 'n' means):

  No, the number of digits of A^2 must be of the form 3n or 3n-1.

Here is my suggested fix for the solution:

It is not possible.  If a number has k digits, then its square will
have 2k or 2k-1 digits.  Therefore, if B has length n, then the total
number of digits used by A and B must be either 3n or 3n-1.

-- Tom Magliery <mag@cs.uiuc.edu>, September 1993

[ Editor: 22 is neither 3n nor 3n-1 for any integer n. ]