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Theorem: Numbers $11,111,1111, \ldots$ are called repunits. No such number is a square.

All repunits are odd numbers. Therefore they must be of the form $4n+1$ or $4n+3$. And since all squares are of the form $4n+1$, one can show that no repunit is a square by showing that all repunits are of the form $4n+3$. It is this claim I would like to prove.

Repunits have the form $10^n+10^{n-1}+\ldots + 10^0$. Summing this as a geometric series one gets $\frac{10^{n+1}-1}{9}$.

The claim is that for all $n \in \mathbb{N} $ there exist $k \in \mathbb{Z}$ such that $\frac{10^{n+1}-1}{9} = 4k+3$. This can be rewritten as $10^{n+1}-28 = 36k$. Assuming $n > 1$ (case $n=1$ can be verified seperately) we get $10^2 \times 10^m - 28 = 36k$ (where $m = n-1$).

Dividing the equation by 4, we get $$25 \times 10^m- 7 = 9k.$$ Rewriting this as a congruence we have $25 \times 10^m \equiv 7 \mod{9}$.

Since $4$ is a multiplicative inverse of $25 \mod{9}$, we have $10^m \equiv 28 \mod{9}$.

So $10^m \equiv 1 \mod{9}$, which is easily verified using the power rule for congruences.

Some teachers probably wouldn't be too happy about this proof, since it is "logically backwards", but that aside, is this proof sound?

Adam
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    An easier approach would be to notice that each of $10^k$ for $k\geq 2$ are divisible by $4$ so is their sum so you're left to prove that $11$ is of the form $4t+3$. – kingW3 Oct 08 '17 at 15:41
  • "Some teachers probably wouldn't be too happy about this proof, since it is "logically backwards", but that aside, is this proof sound?" Um, Other than the fact that it is it not sound is the proof sound? A "logically backwards" proof is never sound. You assume what you need to prove and didn't get a contradiction. That isn't proof. That's a confirmation. You didn't prove that are rep unit must be 4k+3. You just failed to prove that it can't be. But I imagine the proof could be made sound if you can "if and only if" a stateent or two. – fleablood Oct 08 '17 at 15:52
  • Given that you are trying to show that $k$ is an integer, you should probably tie that back in at the end of the proof. – Joffan Oct 08 '17 at 15:52
  • Consider: $10^m + .... + 10 + 1 = 10^2(10^{m-2}+...+1) + 11 = 4(25(10^{m-2}+...+1) + 2) + 3 = 4k+3$. The exact same idea as yours but in only 90 characters. – fleablood Oct 08 '17 at 15:54
  • The title is not correct, as repunits can have different base than 10. In base 3, the repunit 11111 = 1 + 3 + 9 + 27 + 81 is a square. – enrique Jun 11 '19 at 11:11

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Why not just say if $1111.... = 4n+1$ then $4n$ should divide $11111...10$, but this number is not divisible by $4$ using divisibility test of $4$( last 2 digits should be divisible by $4$).

Raghukul Raman
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