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Given that the natural numbers $a, b, c$ are formed by the same $n$ digits $x$, $n$ digits $y$, and $2n$ digits $z$ respectively. For any $n \geq 2$, find the digits $x, y, z$ such that $a^2 + b = c$

Greetings, I was doing this question above and I couldn't figure out how to do.

Here's my progress so far:

The relation $a^2 + b = c$ means $$(xxx...x)^2 + (yy..yy) = (zzz...zz)$$ where the $x\text{ and }y$ are $n$ in number and the $z$ is in $2n$ in number. This simplifies to $$x^2(11···1)^2 +y(11···1) = z(11···1)$$ and I couldn't figure out further. :(

By some trial and error, I found $(333)^2 + 222 = 111,111$.

Any help would be appreciated. I tried but couldn't find out even a similar question somewhere else. Please inform if this question has been asked before.

Thank You

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Let

$$f(n) = \sum_{i=0}^{n-1} 10^i \tag{1}\label{eq1}$$

Thus, as you've shown,

$$x^2 f^2(n) + yf(n) = zf(2n) \implies f(n)\left(x^2 f(n) + y\right) = zf(2n) \tag{2}\label{eq2}$$

Also, note that

$$f(2n) = 10^n f(n) + f(n) = f(n)\left(10^n + 1\right) \tag{3}\label{eq3}$$

Thus, substituting that into \eqref{eq2} and dividing by the common factor of $f(n)$ gives

$$x^2f(n) + y = z(10^n + 1) \tag{4}\label{eq4}$$

First, there's the trivial case of $x = y = z = 0$. The rest of this solution will assume that $z \gt 0$. Next, note that

$$f(n) = \frac{10^n - 1}{9} \tag{5}\label{eq5}$$

Thus, substituting \eqref{eq5} into \eqref{eq4} gives

$$\begin{equation}\begin{aligned} x^2\left(\frac{10^n - 1}{9}\right) + y & = z(10^n + 1) \\ x^2(10^n - 1) + 9y & = 9z(10^n + 1) \\ x^2(10^n) - x^2 + 9y & = 9z(10^n) + 9z \\ (x^2 - 9z)(10^n) & = x^2 - 9y + 9z \end{aligned}\end{equation}\tag{6}\label{eq6}$$

The RHS is less than $100$ (update: actually, it's less than $1000$, but it can be $100$; see the end for more details) and greater than $-10$, so for $n \ge 2$, this means that $x^2 - 9z = x^2 - 9y + 9z = 0$. Note $x^2 - 9z = 0$ occurs only for $x = 3, z = 1$ and $x = 6, z = 4$. This gives, from $x^2 - 9y + 9z = 0$, that $y = 2$ for the first part and $y = 8$ for the second part.

Another way to see this is that the RHS of \eqref{eq4}, as $z$ is a digit, in base $10$, would be $z$, followed by $n - 1$ zeros, and then $z$ again. On the LHS, $y$ is just a single digit. This means for $n \ge 2$, the digits in $x^2f(n)$ must basically all "disappear" when $y$ is added, i.e., $x^2f(n)$ must be just slightly less (i.e., $\lt 9$) than a power of $10$. Checking the squares of $x$ multiplied by $11$, you can see this only occurs if $x = 3$, so $x^2 = 9$ (with it being $1$ less than a power of $10$), or $x = 6$ so $x^2 = 36$ (with it being $4$ less than a power of $10$). For $x = 3$, this means $z = 1$, which requires that $y = 2$, while for $x = 6$, this means $z = 4$ and $y = 8$.

Note the second solution comes from the first one by multiplying by $4$ in \eqref{eq4}. You can see this changes $x = 3$ to $x = 6$, $y = 2$ to $y = 8$ and $z = 1$ to $z = 4$.

In conclusion, there are $3$ solutions, with the sets of digits $(x,y,z)$ being $(0,0,0)$, $(3,2,1)$ and $(6,8,4)$.

Update: As I saw from the answer by fleablood, I made a mistake in my handling of \eqref{eq6}. The RHS of it is definitely less than $1000$, so what I wrote above is true for $n \ge 3$. However, the RHS can be equal to $100$, which means that for $n = 2$, there's also the solution $(8,3,7)$.

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  • $\begingroup$ Great Answer! +1. However, there is one more solution I found that didn't make it in the question. It is $666^2 + 888 = 444,444$ $\endgroup$ – Vasu090 Aug 4 '19 at 5:26
  • $\begingroup$ @Vasu090 Thanks. However, as you pointed out & I realized later, my solution is the only one for larger values of $n$. I'm working on determining & proving it for all the various cases. Please be a bit patient with me while I work on solving this fully. Thanks. $\endgroup$ – John Omielan Aug 4 '19 at 5:31
  • $\begingroup$ OK. No problem John. Thank you for taking the time to answer my question. Really appreciate it. :) $\endgroup$ – Vasu090 Aug 4 '19 at 5:33
  • $\begingroup$ @Vasu090 I realized my mistake and have corrected it. Sorry for not seeing it sooner. As you can see, there's actually $2$ sets of digits $x,y,z$ which work. The second solution includes $66^2 + 88 = 4444$, your value for $n = 3$, $6666^2 + 8888 = 4444 4444$, etc. $\endgroup$ – John Omielan Aug 4 '19 at 5:50
  • $\begingroup$ @Vasu090 I determined there's a simpler, more direct way to determine the solutions. I've added this to my answer, but also left the original arguments as well. $\endgroup$ – John Omielan Aug 4 '19 at 6:15
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If we let $1_n= \underbrace{1111....1}_{n}$

$a^2 + b = c$

$\frac {a^2}{1_n} + \frac {b}{1_n} = \frac {c}{1_n}$

$x^2*1_n + y = z*(10^n + 1)$

Let's think about what that means:

If $x^2 = 10j + k$ we have

$10j + k + y \equiv z$. And we carry $j$ or $j+1$

Then we have $j +k (+1)\equiv 0$. Which means $j+k(+1) = 10$.

This can occur if:

$x=1$ so $k=1$ and $j=0$, $j+k(+1) \equiv 1,2 \not \equiv 0$.

$x = 2$ so $k=4$ and $j=0$, $j+k(+1) \equiv 4,5\not equiv 0$.

$x=3$ so $k=9$ and $j=0$ and $j+k(+1)\equiv 0$ if we carry.

$x=4$ so $k=6$ and $j=1$ and $j+k(+1)\not \equiv 0$.

$x=5$ so $k=5$ and $j=2$ and $j+k(+1)\not \equiv 0$.

$x=6$ so $k=6$ and $j=3$ and $j+k+1\equiv 0$.

$x=7$ so $k=9$ and $j=4$ and $j+k(+1) \not \equiv 0$.

$x=8$ so $k =4$ and $j=6$ and $j+k\equiv 0$.

$x=9$ so $k =1$ and $j = 8$ and $j+k(+1)\equiv 0$.

Now we repeat this and carry again to get the third digit. Assume $n > 2$ As $j+k(+1) = 10$ we must carry $1$ and get $j+k+1$ so this can only happen if $j+k+1 = 10$; not $j+k=10$.

So this can occur if:

$x=3; k=9;j=0$ and $z= j+1=1$ and $9+y \equiv 1$ so $y = 2$.

$x=6; k=9;j=3$ and $z = j+1 = 4$ and $36+y\equiv 4$ so $y=8$.

$x=9; k=1; j=8$ and $z = j+1=9$ and $81+y\equiv 9$ so $y=8$ but.. then we don't carry the $1$. we must have $y=18$ but that's not a single digit.

But if $n = 2$ we can have:

$x^2 = 10j + k$ and $x^2*11 = 100j + 10(k+j) + k$ and $x^2*11 + y= 100j + 10(k+j) + k+y$ where $k+y = z$ and $k+j = 10$ and $j+1 = z$

Then we can have $x=8;k=4;j=6;z=7$ and $y = 3$. (i.e $8^2*11 + 3= 707$ or $(88)^2 + 33 = 7777$

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  • $\begingroup$ I see from your answer that I made a mistake in my answer. I've updated my answer to reference yours and explain what I did wrong. Thanks for your providing your answer here. $\endgroup$ – John Omielan Aug 4 '19 at 7:56
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Please note that

$$\underbrace{1\ldots 1}_{2n}= \underbrace{1\ldots 1}_{n}\underbrace{1\ldots 1}_{n}= \underbrace{1\ldots 1}_{n}\underbrace{0\ldots 0}_{n}+\underbrace{1\ldots 1}_{n}= \underbrace{1\ldots 1}_{n}\cdot1\underbrace{0\ldots 0}_{n}+\underbrace{1\ldots 1}_{n}= \underbrace{1\ldots 1}_{n}\cdot(9\cdot \underbrace{1\ldots 1}_{n}+1)+\underbrace{1\ldots 1}_{n}$$

So if we set $t=\underbrace{1\ldots1}_n$ we can rewrite the equation

$$x^2\cdot \underbrace{1\ldots 1}_{n}^2+y\cdot \underbrace{1\ldots 1}_{n}=z\cdot \underbrace{1\ldots 1}_{2n}$$ as $$xt^2+yt=z(t(9t+1)+t)$$

This simplifies to $$t(9z-x^2)+(2z-y)=0 \tag{1}$$

Now we have two cases:

Case 1: $9z-x^2\ne0$:

Then we have

$$t=\frac{2z-y}{9z-x^2}$$

Because of $x,y,z \in \{0,\ldots,9\}$ and $9z-x^2\ne0$ we have $|2z-y|\le18$ and $|9z-x^2|\ge1.$ So $|t|\le18$, and therefore $t=11$.

If we substitute this in $(1)$ we get

$$11x^2+2y=101z$$

Taking this equation $\pmod {11}$ we get

$$y\equiv 2z \pmod{11} $$

and we can calulate $y$ for every digit $z$. For $z=5$ there is no valid $y$, because $y=10$ is not a digit. From $z$ and $y$ we can calculate $x$. For $z=2$ and $y=4$ there is no valid $x$ because the calculated value is a non integer square root.

z y x
0 0 0
1 2 3
2 4 *
3 6 *
4 8 6
5 * *
6 1 *
7 3 8
8 5 *
9 7 *

Case 2: $9z-x^2=0$

Then we also have

$$2z-y=0$$

The number $9z$ must be a perfect square and therefore $z$ must be a perfect square. So we can calculate

z x y
0 0 0
1 3 2
4 6 8
9 9 *

Because $9z-x^2=2z-y=0$ the equation holds for arbitrary $t.$

So we can conclude that

$$88^2+33=7777$$

and for every $n\ge2$

$${\underbrace{3\ldots3}_{n}}^2+\underbrace{2\ldots 2}_{n}=\underbrace{1\ldots1}_{2n}$$ $${\underbrace{6\ldots6}_{n}}^2+\underbrace{8\ldots 8}_{n}=\underbrace{4\ldots4}_{2n}$$

These are the only solutions.

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