# Can $1\underbrace{44\cdots4}_{n\text{ times}}$ be a perfect power in other bases?

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?

• 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. Dec 6, 2023 at 17:55
• Neither is there another perfect power for the base $5\le b\le 10^5$ and length $4\le l\le 100$ Dec 6, 2023 at 18:05
• @Peter No squares are $4444$ modulo $10000$, as then there would be a square that's $11$ modulo $100$, so $3$ modulo $4$.
– J.G.
Dec 26, 2023 at 10:39

## 2 Answers

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.

(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

https://www.cambridge.org/core/services/aop-cambridge-core/content/view/FC63F15647E84273C77B910D003C0192/S0017089500031207a.pdf/perfect_pell_powers.pdf

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

$$2^{n-1}j^{n+1}+3\cdot2^{n-2}j^n-1=(2j-1)A^2$$

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.

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