# proving that a number is a perfect square

I need help with this problem:

$K$ is equal to $111\ldots111 - 22\ldots22$ where $1$ is used $2n$ times and $2$ is used $n$ times. Prove that $K$ is a perfect square.

It is true, but I can't find a way to prove it. The only interesting thing I've found is that when $n$ is $1, K=3^2$, when $n$ is $2, K=33^2$ and so on. I tried to prove it using induction but wasn't able to find a way to do it, so your help would be really appreciated.

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Write $$111\ldots111 = \sum_{k=0}^{2n-1} 10^k = \frac{10^{2n} - 1}{9}$$ and $$22\ldots22 = 2\sum_{k=0}^{n-1}10^k = \frac{2 \cdot 10^n - 2}{9}.$$

Then $$K = \frac{10^{2n} - 2 \cdot 10^{n} + 1}{9} = \left(\frac{10^{n} - 1}{3}\right)^2.$$ Note that $10^n - 1$ is divisible by $3$, since $10^n \equiv 1^n \equiv 1 (\bmod 3)$.

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Thank you for the answer! –  suomynona Nov 18 '12 at 15:10
You also need to show, or at least mention, that $10^n-1$ is a multiple of 3. –  MJD Nov 18 '12 at 15:23
@MJD Just multiply the first equation by $3$ –  Cocopuffs Nov 18 '12 at 15:37
@MJD I disagree: if $x$ is rational and satisfies a monic equation like $T^2 - x^2 = 0$ then it has to be an integer anyway. I will add it to the answer, though. –  Cocopuffs Nov 18 '12 at 15:50
Since each term was gotten by summing integers, the difference has to be an integer, even though it might not be obvious when looking at the expression. –  marty cohen Nov 18 '12 at 16:10

The number that you are looking at is $$N = (10^{2n-1}+10^{2n-2}+\ldots+10+1)-2(10^{n-1}+10^{n-2}+\ldots+10+1)$$ Summing the GPs, $$N = \frac{10^{2n}-1}{9}-2\frac{10^n-1}{9} = \frac{10^{2n}-2\times10^n +1}{9} = \bigg(\frac{10^n-1}{3}\bigg)^2$$

$\frac{10^n-1}{3}$ will always be of the form $333\ldots$ So, $N$ is a perfect square for all $n$.

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