Can $1\underbrace{44\cdots4}_{n\text{ times}}$ be a perfect power in other bases?
Inspired from this question , I know that $144$ and $1444$ are the only perfect powers in base ten, but in other bases, are there any perfect powers that is form of $1\underbrace{44\cdots4}_{n\text{ times}}$?
I think that the base must be $\geq5$, since $1\underbrace{44\cdots4}_{n\text{ times}}$ contains the number four.
For instance, $144$ is a perfect square in every base, but I think that $1444$ is only a perfect square in base ten.
By using a base converter, $1444$ is not a perfect square in bases that are $3\pmod{4}$, since if $b\equiv3\pmod{4}$, then $1444_{b}\equiv3\pmod{4}$, and all perfect squares are $0, 1\pmod{4}$.
Also, based on this thread, there is generalization that a serd ending with $4444$ (i.e the no. of $4’s$ is more than $4$) cannot be a perfect square in bases that are $2\pmod{4}$.
I know that $1\underbrace{44\cdots4}_{n\text{ times}}$ cannot be a perfect cube or any higher prime power on even bases, as $2^{3}\nmid4444$.
So the only possibilities is that it can be a square in bases that are $1\pmod{4}$, or $2\pmod{4}$(as long as $n=3$ only), and higher odd prime perfect powers in bases that are $3\pmod{4}$.
Take note that the $n$ I’m considering is when $n\geq3$.
By using brute force, I checked the values of $n$ between $4$ up to $10^{4}$, but I was not successful.
Is there a base $b$(other than ten) where $1\underbrace{44\cdots4}_{n\text{ times}}$(where $n\geq3$) is a perfect power?
Best Answer
In the case of $14444$ base $b\ge5$ there can be no squares. We have
$14400_b<14444_b<14641_b$
$(b^2+2b)^2<14444_b<(b^2+2b+1)^2$
and the left and right members are consecutive squares.
With six 4's again there must be no squares:
$1442401_b<1444444_b<1444804_b$
$(b^3+2b^2+1)^2<1444444_b<(b^3+2b^2+2)^2$
We can go on like this. Consider the Laurent series
$\sqrt{1.4444..._b}=\sqrt{1+\dfrac{4b^{-1}}{1-b^{-1}}}=1+\sum\limits_{k=1}^\infty a_kb^{-k}$
$a_1=2,a_2=0,a_3=2,a_4=- 2,...;a_k\in\mathbb{Z}\text{ for all } k$
If $a_m$ is followed by $m$ zeroes in the coefficient sequence then
$P(m)=b^m\left(1+\sum\limits_{k=1}^m a_kb^{-k}\right)$
will be an exact square root of $1444..44_b$ with $n=2m$ fours in all bases; but this can happen only for $m\in\{0,1\}$ because the the known limitations in base ten. For larger $m$, $P(m)$ becomes one of two consecutive whole numbers that strictly bracket $\sqrt{1444...44_b}$. For example, $P(3)=b^3+2b+2$ and we saw above that for $n=2×3=6$ the target square root lies strictly between $P(3)-1$ and $P(3)$. Thereby there are no squares in any base with an even number $\ge4$ of fours.