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Prove that no member of the sequence 11, 111, 1111, ... is a perfect square.

I noticed that the first four terms of the sequence (above) are none of them are a perfect square. But I do not know what to do afterwards.

Username Unknown
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  • Has been done before:http://math.stackexchange.com/questions/298234/prove-that-none-of-11-111-1111-dots-is-the-perfect-square-of-an-intege/298238#298238 – Shaswata Jun 12 '13 at 06:49

3 Answers3

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HINT:

As $\underbrace{11\cdots11}_{n\text{ terms }}$ is odd, its square root must be odd

$$\text{Now, }(2a+1)^2=4a^2+4a+1\equiv1\pmod 4$$

$$\text{But, }\underbrace{11\cdots11}_{n\text{ terms }}\equiv11\pmod{100}\equiv-1\pmod 4\text{ for }n>1$$

lab bhattacharjee
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Since this particular question has been discussed before, let me add five coins by proving that each number of the form $\overline{aaaa....aaaa}$ can not be a perfect square. Indeed, suppose that $\overline{aaaa....aaaa}=k^2,$ in other words $$k^2=a\cdot\frac{10^n-1}{9}.$$ Rewriting the last inequality we end up with $$a(10^n-1)=(3k)^2.$$ Considering both sides modulo $5,$ we get that $a=0,1,-1(\mod 5).$ Note, that $a\ne 5,$ because LHS is not divisible by $5.$ Considering both sides modulo $4,$ we have that $a=0,-1 (\mod 4)$ because $n\ge 2.$ Combining last two observations we end up with this only possibility $a=4.$ Then, $k=2k_1$ and we end up with $$10^n-1=(3k_1)^2$$ which is impossible modulo $4$ for $n\ge 2.$

leshik
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  • Unfortunately this fails when one allows a>10. But in general, rep-units that are proper powers is pretty rare, but rep-digits in other bases are not all together rare: eg $3,3,3$ in base 22, and $7,7,7$ in base 18 are both square. So is $30,30,30,30$ in base 239. In base 68, $17,53,17,53,17,53$ is a square, and $20,9,20,9$ is a cube, which is pretty unusual. – wendy.krieger Jun 12 '13 at 10:00
  • I understand what you are doing up to the point were you are considering the different cases for a. Can you say a=1, and then manipulate the inequality? – Username Unknown Jun 12 '13 at 17:59
  • well, we have to consider cases $a=1,2,3...9,$ not just $a=1.$ To rule out some of them, we work modulo $4$ and $5,$ to get those congruences. The only number that satisfies both congruences is $a=4.$ – leshik Jun 12 '13 at 18:36
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There is no proof it holds in general, because in base 3, the numbers $2^2=11$ and $102^2=11111$ are both square, and in base 7, the number $26^2=1111$ is square.

In decimal, the numbers reduces to $3$ mod 4, which no square produces.

One might note also that Shank's prime $487$ is one of two known primes in decimal, that if $p$ divides a rep-unit, then so does $p^2$. The other prime is $56598313$. 3 also divides its period (to complete the picture), but its period is 1, which means that $3^2|10^x-1$ for all x, but not $(10^x-1)/(10-1)$.

wendy.krieger
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