What is the solution to this? I found that I must have $x=y,z=0$ to have more then to solutions. Is that right?

Since the thing's under radical, $y<x$. Thus, we have, $\sqrt{(x-y).111....1}=\underbrace{\overline{zz\cdots z}}_{n\text{ times}}$ Thus no soln. except when $(x-y)=0$ because a number of the form $(x-y).111....1$ cannot be a perfect square. Here's why, $11..1\rightarrow$ is the form of $4k+3$ so cannot be a perfect square.(Except when $n=1$, that is for $1$, but this case is immaterial since the values has to satisfy for at least two values of $n$) So if $gcd(x-y,111...1)=1$ it cannot be a perfect square. So this rules out $(x-y)=1,2,4,5,8$ No perfect square ends with $3$ or $7$ which rules out $(x-y)=3,7$ For $9$, $(x-y)$ comes out of the radical sign and we are left with $\sqrt{111...1}$ ruling out $(x-y)=9$ And for $66..6$, $2|666...6$ but $4$ doesn't divide it. So only soln. when $x=y$ and $z=0$ and this holds good $\forall n \in \mathbb{N}$ Peace. @ myro111: you're right Faustus.

sqrt(7777-33)=88

94337sk wrote: sqrt(7777-33)=88 Please read the question properly. Amir Hossein wrote: Find digits $x, y, z$ such that the equality \[\sqrt{\underbrace{\overline{xx\cdots x}}_{n \text{ times}}-\underbrace{\overline{yy\cdots y}}_{n \text{ times}}}=\underbrace{\overline{zz\cdots z}}_{n \text{ times}}\] holds for at least two values of $n \in \mathbb N$, and in that case find all $n$ for which this equality is true. The number of times digit $x$ appears must be the same as the number of times the digit $y$ appears.

The whole thing depends if in Iran $0\in \mathbb{N}$ or not. If not, the proof that $x=y$, $z=0$, is the unique solution may be drastically simplified over that presented at post #3. Denote $A_n=\overline{\underbrace{11\ldots 1}_{n \textrm{ times}}} > 0$. We must have $(x-y)A_n = z^2A_n^2$, thus $A_n = z^2A_n$, but for $z\neq 0$ and $n>1$ we have $zA_n > 10 > x-y$ for any $x,y$. If however yes, then the equality always holds for $n=0$, and it also holds for $z\neq 0$ and $n=1$ if $x-y = z^2$, possible for $z=1$ ($x=y+1$), $z=2$ ($x=y+4$), or $z=3$ ($x=9$, $y=0$).

YESMAths wrote: 94337sk wrote: sqrt(7777-33)=88 Please read the question properly. Amir Hossein wrote: Find digits $x, y, z$ such that the equality \[\sqrt{\underbrace{\overline{xx\cdots x}}_{n \text{ times}}-\underbrace{\overline{yy\cdots y}}_{n \text{ times}}}=\underbrace{\overline{zz\cdots z}}_{n \text{ times}}\]holds for at least two values of $n \in \mathbb N$, and in that case find all $n$ for which this equality is true. The number of times digit $x$ appears must be the same as the number of times the digit $y$ appears. No he is correct here. The OP made a error. It should be $2n$ times for the digit $x$. see page 37 no.12 http://web.cs.elte.hu/~nagyzoli/compendium.pdf Fixed ~dj

x = 1, y = 2, z = 3 works for any n (this can be seen by writing LS and RS each as a sum and then reducing).

Bump! There isn't any promising solution!

Ig $(x,y,z)=(1,2,3),(4,8,6),(0,0,0)$ works for any $n$. The equation is equivalent to $10^n(z^2-9x)=(z^2+9x-9y)$ Thanks @2below

what about (7,3,8)?

Pluto1708 wrote: Ig $(x,y,z)=(1,2,3),(4,8,6),(0,0,1),(0,0,0)$ works for any $n$. The equation is equivalent to $10^n(z^2-9x)=(z^2+9x-9y)$ (0,0,1) isnt working..

Since, there was no solution to this problem, thought I'd just post mine