Which version of this question is right?

Question

Find digits $x,y,z$ such that the equality $$\sqrt{\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}}} = \overline{\underbrace{zz\cdots z}_{n}}$$ holds for at least two values of $n \in \mathbb{N}$, and in that case find all $n$ for which this equality is true.

Is my method below correct?

Let $A_n=\overline{\underbrace{11\ldots 1}_{n}}$. Then we have $$\sqrt{\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}}} = \sqrt{\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$n$}}\cdot 10^n+xA_n-yA_n}$$ and thus equals $\sqrt{\smash[b]{A_n(x10^n+x-y)}}$.

Lemma. A number of the form $A_n(x10^n+x-y)$ can't be a perfect square.

Proof. We know that $A_n = 4k+3$, which is never a perfect square, except possibly when $n = 1$. Therefore in the case $\gcd(A_n, x10^n+x-y) = 1$, we know that the only time a solution will occur is when $n=1$. So we can disregard this case. Now, if $\gcd(A_n, x10^n+x-y) \neq 1$, that must mean $x=0$, an impossibility. $\square$

JimmyK4542 · Answer 1 · 2016-05-11T01:00:15.390

1

No, unfortunately your solution is not correct.

Your proof breaks down when you make the claim:

Now, if $\gcd(A_n, x10^n+x-y) \neq 1$, that must mean $x=0$, an impossibility.

If say $n = 2$, $x = 1$, $y = 2$, and $z = 3$, then we have $\gcd(A_n, x10^n+x-y) = \gcd(11,99) = 11 \neq 1$, but $x = 1 \neq 0$.

So your claim is false in this case.

Hint for solving the problem: Try writing the given equation as: $$\sqrt{\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}}} = \overline{\underbrace{zz\cdots z}_{n}}$$

$$\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}} = (\overline{\underbrace{zz\cdots z}_{n}})^2$$

$$\dfrac{x(10^{2n}-1)}{9} - \dfrac{y(10^n-1)}{9} = \dfrac{z^2(10^n-1)^2}{9^2}$$

$$9x(10^n+1) - 9y = z^2(10^n-1)$$

$$9x(10^n-1) + (18x-9y) = z^2(10^n-1)$$

$$(9x-z^2)(10^n-1) = 9y-18x$$

Can you figure out values of $n,x,y,z$ that work from this equation? Remember that $x,y,z$ are digits which must be between $0$ and $9$.

edited May 11 '16 at 01:00

answered May 10 '16 at 23:59

JimmyK4542

54,331

How do I find $n,x,y,z$? – user19405892 May 11 '16 at 00:28
^I just edited it in a way that should make it more obvious. Let me know if you are still stuck. – JimmyK4542 May 11 '16 at 01:00
I am still stuck. – user19405892 May 11 '16 at 01:13
^Start with the case where $n \ge 3$. You know that $(9x-z^2)(10^n-1) = 9y-18x$ is between $9 \cdot 9 = 81$ and $-18 \cdot 9 = -162$. But you also know that $(9x-z^2)(10^n-1)$ is an integer multiple of $10^n-1$. What does this tell you? For $n = 1$ and $n = 2$, you can simply find all the solutions by brute force. – JimmyK4542 May 11 '16 at 01:28
I think you mean $n \geq 2$, not $3$ right? – user19405892 May 11 '16 at 01:40
No, I mean $n \ge 3$, There are more solutions for $n = 1$ and $n = 2$ than there are for any fixed value of $n$ that is $3$ or larger. So we need to handle those cases separately. – JimmyK4542 May 11 '16 at 01:46
Can you give an example of values for $x,y,z,n$ where only one of the first and last equations is satisfied? – JimmyK4542 May 11 '16 at 02:18
I see the issue. When you simplified it was under the assumption $n \neq 0$. So what you did is fine. Also, for $n=2$, the LHS can't possibly be anything in the range $-162$ to $81$ so that's why I am saying it doesn't work. – user19405892 May 11 '16 at 02:20
Let us continue this discussion in chat. – JimmyK4542 May 11 '16 at 02:26

Which version of this question is right?

1 Answers1