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?

  • 1
    $\begingroup$ A further perfect power in base $10$ would have to exceed $10^{3\cdot 10^5}$ , for me enough to be convinced that there is no further. $\endgroup$
    – Peter
    Dec 6, 2023 at 17:55
  • 1
    $\begingroup$ Neither is there another perfect power for the base $5\le b\le 10^5$ and length $4\le l\le 100$ $\endgroup$
    – Peter
    Dec 6, 2023 at 18:05
  • 1
    $\begingroup$ @Peter No squares are $4444$ modulo $10000$, as then there would be a square that's $11$ modulo $100$, so $3$ modulo $4$. $\endgroup$
    – J.G.
    Dec 26, 2023 at 10:39

2 Answers 2


In the case of $14444$ base $b\ge5$ there can be no squares. We have



and the left and right members are consecutive squares.

With six 4's again there must be no squares:



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.


(I'll try make updates on this when I find something else)

hopefully I'm not wrong :\

Let the perfect power be $a^m$.

We can see that when $n=1$, $m=2$ will always work iff $b+4$ is a perfect square. ($b \geq 5$ is assumed throughout here).

When $n=m=2$, all $b \geq 5$ will work, as the result will always be $(b+2)^2$.

From now on, we will only be dealing with $n \geq 3$.

There are no perfect $m$th powers when $b=5$.

The following case is where $m$ is $\textbf{odd}$, and $\textbf{not a multiple of $5$.}$

We can rearrange the question to this form:

$$b^n + 4\frac{b^n-1}{b-1} = \frac{b^{n+1}+3b^n-4}{b-1} = a^m$$

Subbing $b=5$ yields $$\frac{8\cdot5^n-4}{4} = 2\cdot5^n-1 = a^m$$

Thus, $a$ must be odd.

$$2 \cdot 5^n = (a+1)(a^{m-1}-a^{m-2}+\cdots +a^2-a +1)$$

Thus, we know that $2 \mid a+1$. Since the rest are just powers of $5$, as well as $a+1 \leq a^2-a+1$ for $a \geq 2$, we have that

$$a+1 \mid 2(a^{m-1}-a^{m-2}+\cdots +a^2-a +1)$$

However, we also have that

$$a+1 \mid 2(a^{m-1}-a^{m-2}+\cdots +a^2-a -(m-1))$$

Thus, $$a+1 \mid 2\cdot m$$

Since $m$ is odd, $2$ cannot divide $m$. Since $m$ is also not a multiple of $5$, we must have that $$a+1 \mid 2$$

Since $(a+1)$ is also a factor of $2 \times 5^n$, we have that $a+1=2$ and so $a=1$. This gives us that $a \cdot 5^n = 2$ and so $n=0$, which is impossible.

The following case is where $m$ is $\textbf{odd}$, and $\textbf{is a multiple of $5$.}$

Now, if $m \equiv 0 \pmod 5$, and $m$ is odd, then let $m= 5^s \cdot M$, where $M$ is not divisible by $5$. Let ${a^5}^s = A$. Then, we have that $$a^{5m}= a^{5^s \cdot M} = ({a^{5^s}})^M = A^M$$

Again, we have that $$2 \cdot 5^n -1 = A^M$$

We can apply the solution above since $M$ is not a multiple of $5$. This gives us that $A+1=2$, and so $A=1$ and $n=0$, a contradiction.

If $m$ is a perfect power of $5$, then we will rearrange it slightly. Let $a^{5^{s-1}} = A$. Then, we have that $2 \cdot 5^n = a^5 + 1$. If we expand it, and use the above techinque, we get that $a+1 \mid 10$.

$a=9$ is the only option since $a$ has to be odd. $a=1$ leads to a contradiction. However, subbing it in reveals

$$2 \cdot 5^n = 9^5 - 1$$

Clearly, the RHS is divisible by $4$, but the LHS isn't, so this is a contradiction.

The following case is where $m$ is $\textbf{even}$.

We have that $m=2^r \cdot J$.

Like above, let $\alpha = {a^2}^r$. Then we have that

$$2 \cdot 5^n -1 = a^m = a^{2^r \cdot J} = (a^{2^r})^J = \alpha^J$$

Here, $J$ is odd, and so depending on whether $J$ is divisible by $5$, it can be dealt by the previous cases, unless $J=1$.

Assuming that $m \neq 2$, let $t = a^{2^{r-2}}$. We then have that $ 2\cdot 5^n - 1 = t^4$.

Multiplying both sides by $4$, we get

$$8 \cdot 5^n - 4 = 4t^4$$

Adding 16 to both sides yields

$$8 \cdot 5^n + 12 = 4t^4 + 2^4$$

Using Sophie Germain's identity https://en.wikipedia.org/wiki/Sophie_Germain%27s_identity :

$$8 \cdot 5^n + 12 = (2t^2+4t+4)(2t^2-4t+4)$$

Dividing by $4$ yields

$$2 \cdot 5^n + 3 = (t^2+2t+2)(t^2-2t+2)$$

Now, let's check all possible cases for $t$ modulo $5$.

$t \equiv 0$ yields RHS as $4$.

$t\equiv 1,2$ yields RHS as $0$ (First term $= 0$)

$t\equiv 3,4$ yields RHS as $0$ (Second term $= 0$)

Note that the LHS is $3$. Thus, there are no solutions.

Thus, we are left with only the $m=2$ case. The first lemma in this paper by John Cohn


states that all solutions to the Diophantine equation $2z^k - 1= y^2$ where $k>2$ are $y=z=1$ and $y=239,z=13,k=4$. However, with our base being $5$, $z=5$ is not part of a solution and thus there are no squares.

There are no perfect $m$th powers when $b = 2k$, for $m \geq 3$.

Let $b=2 \cdot k$. We get that $$(2k)^{n+1}+3(2k)^n-4=(2k-1)a^m$$ $$2^{n+1}k^{n+1}+3\cdot2^nk^n-4=(2k-1)a^m$$ Since $n \geq 3$, we can see that the LHS can be factored to form $$4(2^{n-1}k^{n+1}+3\cdot2^{n-2}k^n-1)=(2k-1)a^m$$ Notice that the LHS is even, and $2k-1$ is odd, so $a^m$ must be even, and thus $a$ must be even. However, there are no more than 2 powers of $2$ on the LHS, so if $m \geq 3$, there will not be enough powers of $2$.

Thus, we are done.

There are no perfect $m$th powers when $v_2(b)=1,2$ unless $n=3$.

It suffices to show that there are no squares, from the result above.

Now, if $n \geq 4$ and $v_2(b)=1$:

Rewrite $b$ as $2j$.

We yield that $$2^{n+1}j^{n+1}+3\cdot2^nj^n-4=(2j-1)a^2$$


where $a=2A$.

Now, note that $j$ is odd here. Thus, $2j -1 \equiv 1 \pmod 4$.

Now, taking $\pmod 4$ gives us that $-1 \equiv a^2 \pmod 4$ if $n-2 \geq 2$, which is the $n \geq 4$ requirement. However, this cannot happen.

This argument can be repeated similarly for $\pmod 8$, which leaves us with the case of $n=4$. Then, according to @Oscar_Lanzi's result, this contains no squares, and thus we are done.

  • $\begingroup$ This should be "all solutions to the Diophantine equation with $k>2$". There are an infinite number of solutions with $k=2$: for example $z=5$, $y=7$ . $\endgroup$
    – mcd
    Dec 29, 2023 at 8:25
  • $\begingroup$ oh sorry ill go fix it $\endgroup$ Dec 29, 2023 at 8:29
  • $\begingroup$ It just gives the $12^2=144$ case in base 5. $\endgroup$
    – mcd
    Dec 29, 2023 at 8:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .