## Task $6 / 76$

Find the base $x$ of the number system in which the equation holds: $(211)_{x}^{2}=(100021)_{x}$.
By definition, we have

$$
(211)_{x}=2 x^{2}+x+1 \quad ; \quad(100021)_{x}=x^{5}+2 x+1
$$

Accordingly, the given equation takes the form

$$
\left(2 x^{2}+x+1\right)^{2}=x^{5}+2 x+1
$$

This equation is equivalent to

$$
x^{2}\left(x^{3}-4 x^{2}-4 x-5\right)=0
$$

Since $x=0$ cannot be a base of a number system, we need to investigate the integer solutions of the equation

$$
x^{3}-4 x^{2}-4 x-5=0
$$

Since $x\left|x^{3}, x\right| 4 x, x|4 x, x| 0$, it follows that $x \mid 5$. The divisors $\pm 1$ are also not valid for a base of a number system, leaving only $\pm 5$. Testing these, only the number +5 is confirmed. By polynomial division of $x^{3}-4 x^{2}-4 x-5$ by $x-5$, we obtain the polynomial $x^{2}+x+1$, which has no (real) roots. Thus, the only solution is $x=5$. Therefore, the equation $(211)_{5}^{2}=(100021)_{5}$ holds.

## Solution by Rosita Haase:

The product 211 * 211 should yield 100021 in the considered number system. Using the "conventional" method, we get

$211 * 211$

44521

Comparing 44521 with 100021, we notice that $2+1+2=5=0+$ carry to the tens place. Therefore, the base must be 5.