444444....4444888888....8889

[Where there are 'k' Fours, '(k-1)' Eights and 'Exactly One' 9],

are always perfect squares.

(For example the sequence of numbers: 49, 4489, 444889, ....etc. and so on are always perfect squares).

Ignore the 9 to start with. It is a serious distraction.

4/9*(10^n-1)*10^n gives the 4's. 8/9*(10^n-1) gives the 8's.

Adding 1: 4/9*(10^n-1)*10^n+8/9*(10^n-1)+1=1/9 (2*10^n+1)^2. 

But (2*10^n+1) is simply 201,2001,20001, etc, always divisible by 3.

So its square will always be divisible by 9; 1/9(3n)^2=n^2.

(The 'official solution' dances all around this point, without ever addressing its consequence.)

Couldn't be easier. Well worth a vote though!

Edited on September 17, 2012, 3:26 am

Posted by broll on 2012-09-17 03:10:19