# Prove the digital root of a square can be anything other than $2, 3, 5, 6, 8$?

The digital root is the sum of the digits, unless that has more than one digit, so then you add up the digits again, until arriving at a single digit, e.g., $28$ -> $2 + 8 = 10$ -> $1 + 0 = 1$.

For what the digital root of a square can be, I just need to give an example of it, right? $0, 1, 4, 9$ are easy enough, and for $7$ I can just do $16$ or $25$ or $169$ or $196$ or $484$ or $529$ etc. (they seem to follow each other like that for some reason).

I'm not a professional mathematician, so please forgive me if what I'm saying is too obvious or too easy. I have looked among the first thousand squares and failed to find a single one with digital root $2, 3, 5, 6$ or $8$, but that doesn't prove anything, because maybe $1001^2$ does something contrary to expectation. I have this feeling that the explanation is in plain sight but for some reason I'm failing to see it.

• The digital root of a number is its remainder modulo $9$ (where remainder $0$ becomes digital root $9$, no biggie). So the question is which remainders modulo $9$ a square can have. Do you know a little modular arithmetic? Jun 17, 2015 at 20:43
• Only a little, I will have to review, very rusty.
– Bob
Jun 17, 2015 at 20:49

The “digital root” is obtained by casting out nines. The arithmetic basis of this is that we remove multiples of $9$, so, if $d(n)$ denotes the digital root of $n$, we have $$n\equiv d(n)\pmod{9}$$ meaning that $n-d(n)$ is a multiple of $9$.

Now, a basic property of congruences (that is, relations like the one above) is that they respect addition and multiplication. Thus $$d(n^2)\equiv n^2\equiv (d(n))^2\pmod{9}$$ So we just need to compute the digital root of the squares of digits: $$d(1^2)=1 \quad d(2^2)=4 \quad d(3^2)=9 \quad d(4^2)=7 \\ d(5^2)=7 \quad d(6^2)=9 \quad d(7^2)=4 \quad d(8^2)=1$$

First of all, $1001^2$ is not going to break the pattern you've observed so far, being $1002001$ and thus having a digital root of $4$. Squares with a digital root of $4$ follow or precede squares with a digital root of $1$, like $1000^2$.

But you're right in general to distrust the evidence of a thousand examples. In this particular case, the important thing to remember is that a problem of base $10$ digital roots is essentially a problem in arithmetic modulo $9$ (with just a couple of caveats to keep in mind), and modular arithmetic is periodic. Another neat feature of congruences is that you can multiply them and the results are valid, e.g., $2 \times 4 = 8$, $2 \times 5 = 1$, $2 \times 6 = 3$, etc.

So instead of trying to look at the infinitely many squares, you just need to look at the squares of $1, 2, 3, 4, 5, 6, 7, 8, (9)$, to get $1, 4, (9), 7, 7, (9), 4, 1, (9)$ (this also explains why the squares with digital root $7$ "seem to follow each other like that"). For the most part so far I've merely restated what others have said.

There is a slightly different way you can go about proving neither $3$ nor $6$ can be the digital root of a square in base $10$. In base $10$, we have these divisibility tests for $3$ and $9$: if the digital root if $3$, $6$ or $9$, then the number is divisible by $3$, and if the digital root is $9$, then the number is divisible by $9$. Then, if $n$ is a nonzero integer, then $3n$ has a digital root of $3$, $6$ or $9$. But $(3n)^2 = 9n^2$ and therefore it must have a digital root of $9$. So if a number $m$ has $3$ or $6$ for a digital root, that means it's divisible by $3$ but not by $9$, and $\sqrt{m} = x \sqrt{3}$. But $\sqrt{3}$ is not an integer. The only drawback of this method is that you can't easily extend it for $2, 5, 8$.

Think about these $\bmod 3$.

• If $n = 3k$, then $n^2 = 9k^2$.
• If $n = 3k + 1$, then $n^2 = 9k^2 + 6k + 1$.
• If $n = 3k$, then $n^2 = 9k^2 + 12k + 4 = 9k^2 + 12k + 3 + 1$.

Therefore $n^2 \equiv 2 \pmod 3$ is impossible if $n$ is an integer. What dose this have to do with digital roots? Well, if $n^2 \equiv 2, 5, \textrm{ or } 8 \pmod 9$, then $n^2 \equiv 2 \pmod 3$. But as I just demonstrated above, that congruence has no solutions.

To prove neither $3$ nor $6$ can be the digital root of a square, we have to broaden back out to $\bmod 9$. If $n = 9k + 3$, then $n^2 = 81k^2 + 54k + 9$, which is obviously divisible by $9$. Or if $n = 9k + 6$, then $n^2 = 81k^2 + 108k + 36$, which, again, is obviously divisible by $9$.

The digital root is actually equal to the number $\mod 9$ (the remainder when divided by 9). The only difference is that $0$ is replaced by $9$, except of course when the number is $0$.

Looking at the squares $\mod 9$, we see 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, 4 ....

Therefore, this pattern begins to repeat. And we see that the unattainable values are $2, 3, 5, 6, 8$.

The digital root of $n^2$ has the same remainder upon division by $9$ as $n^2$ itself and is uniquely determined by this /except that "divisible by $9$" gives digital root either $9$ or $0$). The remainder of $n^2$ again depends only on the remainder of $n$. So checking $n=0,1,2,3,4,5,6,7,8,9$ is enough to find all possible values.