Digital Roots of Square Numbers Can anyone offer a proof of the following: 
The digital root of a square number is always $1$, $4$, $7$ or $9$. (It is never $2$, $3$, $5$, $6$ or $8$.) 
Digital root : Add the digits of a number until you get a single digit. 
examples: The digital root of $144$ is $1+4+4 = 9$. 
The digital root of $14289$ is $1+4+2+8+9 = 24  2+4 = 6$. 
The digital root of $1428842$ is $1+4+2+8+8+4+2 =  29  2+9 = 11 1+1 = 2$. 
Square number: A number whose square root is an integer. 
Examples: $25$    $36$    $144$    $400$ 
$116$ is not a square number. 
 A: Mod $9$: $0^2 \equiv 3^2 \equiv 6^2 \equiv 0$, $1^2 \equiv 8^2 \equiv 1$, $2^2 \equiv 7^2 \equiv 4$, $4^2 \equiv 5^2 \equiv 7$.  That's all!
A: Explained in some more detail. The procedure for taking the digital root preserves the remainder modulo 9. Also, the digital root is a single digit i.e. it's 0 to 9 (btw, it's 0 just for the 0 itself). So if your initial number has a remainder of e.g. 6, then the digital root will also have remainder of 6 (modulo 9), which actually means it will be equal to 6 (as it's a single digit). Now look at the possible remainders which a square number can have modulo 9. They are just 0, 1, 4, 7 (never 2,3,5,6,8). So the digital root will also have a remainder of 0, 1, 4, or 7 (modulo 9), which means it will be equal to 9, 1, 4, or 7. 
A: To  understand this phenomena, you should read Modular arithmetic. There are you will find general ideas.
A: Taking the digital root is equivalent to taking the remainder of a number divided by 9.
So we'll want to look at,
$$ n \cdot n \equiv v \mod 9$$
Where, $v=1,4,7,9$
This simplifies to,
$$ (n \mod 9)^2 \equiv v \mod 9$$
Of course $n \mod 9$ can only be inclusively between $0$ and $9$. We're done, simply enumerate all the values from $0$ to $9$ and verify the above, it's clear that they satisfy the conditions for v. 
