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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.

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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.

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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!

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  • $\begingroup$ @mathlove Thanks for catching that. $\endgroup$ – Robert Israel Jul 28 '15 at 22:39
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To understand this phenomena, you should read Modular arithmetic. There are you will find general ideas.

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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.

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