Are there an infinite number of squares that do not contain the digit 0?
Prove your assertion or negation.
All we need is some definable sequence of perfect squares to satisfy the problem statement. So I found the sequence 49, 4489, 444889, 44448889, .... These are the squares of 7, 67, 667, 6667, ....
The base numbers can be written as 666..666+1 = (2/3)*(10^n-1)+1
Square this to get (4/9)*(10^n-1)^2 + (4/3)*(10^n-1) + 1
= (4/9)*(10^(2n)-1) - (8/9)*10^n + 8/9 + (4/3)*(10^n) - 4/3 + 1
= (4/9)*(10^(2n)-1) + (4/9)*(10^n-1) + 1
This last expression can be expanded into 4444...4444 + 88..88 + 1, which is equal to our sequence of squares.
Solution
