Is there a number system base $a$ where $2018$ can be written as $\overline{21312}^{a}$ in it?

Question

That's actually the whole question.

Is there a number system base $a$ where $2018$ can be written as $\overline{21312}^{a}$ in it?

I honestly don't know what I do more than this

$2a^4 + a^3 + 3a^2 + a = 2016$

I don't know how to solve an equation like this and how to check if it has solutions.

EDIT: It seems that the book forgot to mention it in the question but $a > 3$

Answer 1

$$2a^4 + a^3 + 3a^2 + a + 2 = 2018$$ Since $a$ is positive: $$2a^4 < 2016$$ $$a^4 < 1008$$ $$a < 5.6$$ Since $a$ is integer: $$a <= 5$$ Now let's take $a = 5$: $$2a^4 + a^3 + 3a^2 + a + 2 = 1452$$ If we take $a < 5$, we will get numbers less than 1452.

Means, there is no solution.

Answer 2

Observe that if you write $2018$ in base $a$ as $\overline{21312}^a$, then it leads to the following polynomial $2a^4+a^3+3a^2+a+2=2018$. Equivalently we have that we want an integer root of the polynomial $2a^4+a^3+3a^2+a-2016$. Now we can use the rational roots theorem to just check the divisors of $2016$. That is we must check if $a=1,2,3,4,6,7,8,9,12,14,16,18,21,24,28,32,36,42,48,56,63,72,84,96,112,126,144,168,224,252,288,336,504,672,1008,2016$ are roots. Of which after checking them all none of them are roots.

Another solution would be to note that $a>3$, and just check $a=4,5,6,7$, and note that when $a=4$, $\overline{21312}^a=630$, when $a=5$ we have $\overline{21312}^a=1457$ when $a=6$ we have that $\overline{21312}^a=2924$, and as this polynomial is increasing for $a$ positive, we conclude that there is no solution.

Answer 3

Well as $2131$ has more digits than $2018$, $a$ must be a base less than $10$.

And as $a$ divides $2a^4+a^3 + 3a^2 + a$ we must have $a| 2016= 2^5\cdot 3^2\times 7$ we must have only have $2,3,7$ as prime divisors. So $a=2,4,8,3,6,9,7$.

Also $2016\div a = 2a^3 + a^2+3a + 1$. But as $2^5|2016$ but none of $2,4,8,3,6,9,7$ have a power as high as $2^5$ dividing them, so $2016\div a$ must be even. So $2a^3 + a^2+3a +1$ must be even. But if $a$ is even then $2a^3+a^2 + 3a +1$ is odd and if $a$ is odd then $2a^3 + a^2 + 3a + 1$ is odd. So no solutions are possible.

Answer 4

$2a^4+a^3+3a^2+a \equiv 2016\pmod 2$
$a^3+a^2+a\equiv 0\pmod 2$
$a(a^2+a+1)\equiv 0\pmod 2$
$a^2+a+1$ is odd.
$a\equiv 0\pmod 2$
$2\mid a\implies 2^2\mid 2a^4+a^3+3a^2,2016\implies 2^2\mid a\implies 2^4\mid 2a^4+a^3+3a^2,2016\implies 2^4\mid a$
$a\geq 16\implies 2a^4+a^3+3a^2+a>2a^4\geq 2*16^4>2*10^4>2016$
contradiction
so $2a^4 + a^3 + 3a^2 + a = 2016$ has no positive integer solutions for $a$.