TITLE: Proving that a natural number made entirely of 6's and 0's is not a square.
QUESTION [11 upvotes]: I proceeded using infinite descent. Let $ N =a_na_{n-1}a_{n-2}\ldots\ldots a_2a_1a_0$ be the decimal representation of the number. Then either $N$ ends in an even number of zeroes  or $a_0=6$
Now all squares are $\equiv 0 \text{ or } 1 \bmod 4 $. But  if $N$ ends in $06\text{ or }66 $, then $N\equiv 2 \bmod 4 $. Thus $ N $ does not end in $6$.
If $N$ ends in an even number of zeroes then, then $ N=10^{2n}\cdot N'$.
Applying the same argument to $ N'$ we commence an infinite descent.
Is the proof correct ? Any shorter proofs ?

REPLY [2 votes]: Let $a_1a_2a_3...00$ be a square number with $2k$ number of zeroes at the end, and the number is just made of $0's$ and $6's$ 
Let us suppose that's a square, $x^2=a_1a_2a_3...00$
$x^2=a_1a_2...00=a_1a_3x_4..a_m.10^{2k}$
Which means $a_1a_3a_4..a_m$ is also a square, since $a_{m-1}a_m=06$ or $66$.
$a_1a_2..a_m= 2\mod 4$. 
Contradiction. Therefore, there's no such square! And yeah. Your argument is right. :)