Proving that a natural number made entirely of 6's and 0's is not a square. 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 ?
 A: To make the last part of the argument of @lsp explict, if a square $x^2$ ends in $6$, then the last two decimal digits of $x$ are


*

*either $a4$, and then
$$
x^2 \equiv (a \cdot 10 + 4)^2 \equiv a \cdot 80 + 16 \pmod{100},
$$

*or $a6$, and then
$$
x^2 \equiv (a \cdot 10 + 6)^2 \equiv a \cdot 120 + 36 \pmod{100}. 
$$


In both cases the last-but-one digit is indeed odd.
A: If $N=10^{2n}.P$ , then $10^{2n}$ is already a perfect square. We just need to check if $P$ is a square. 
Since it can either end with '$06$' or '$66$' this is never a perfect square as perfect squares ending with '$6$' should always have an ODD number in ten's place.
A: 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 $66$.
$a_1a_2..a_m= 2\pmod 4$.
Contradiction. Therefore, there's no such square! And yeah. Your argument is right.
