Can $4\cdots41$ (with odd number of $4$s) be a Square Number?

Question

Consider a number in its decimal representation that begins with an odd number of consecutive digits of 4, followed by a single digit of 1. An example of such a number would be 41, 4441, or any similar pattern extending with 4s. My question is: Can such a number ever be a perfect square?

To clarify, the numbers we're considering take the form 44 … 41 44…41, where the number of 4's is odd, and it's terminated by a single 1.

Here are the specific points I'm curious about:

1. Is there a mathematical approach or theorem that directly addresses the properties of numbers with specific digit patterns in relation to being perfect squares?
2. Could modular arithmetic or any form of number theory provide insight into proving or disproving the possibility of such a number being a perfect square?

I've attempted some preliminary analysis, including playing around with smaller cases and considering the last digits of square numbers, but haven't reached a conclusive answer. Any guidance, references, or insights into how to approach this problem would be greatly appreciated.

Thank you in advance for your time and assistance!

Answer 1

A simpler approach. Number $44…4$, consisting of even number of $4$s is always divisible by $11$. Hence our number $44…41$ with odd number of $4$s will be $8$ modulo $11$. But these are the only possible remainders of squares modulo $11$: $$0^2=0$$ $$(\pm1)^2=1$$ $$(\pm2)^2=4$$ $$(\pm3)^2=9$$ $$(\pm4)^2=5$$ $$(\pm5)^2=3$$ All equalities are modulo $11$.

Answer 2

No. Such a number is always between $66...66^2$ and $66...67^2$.

Here are three families of square numbers that have simple repeating patterns:

$${\underbrace{33...33}_n}^2={\underbrace{11...11}_{n-1}}0{\underbrace{88...88}_{n-1}}9$$ $${\underbrace{66...66}_n}^2={\underbrace{44...44}_{n-1}}3{\underbrace{55...55}_{n-1}}6$$ $${\underbrace{99...99}_n}^2={\underbrace{99...99}_{n-1}}8{\underbrace{00...00}_{n-1}}1$$

The third family is easy to prove by writing $99...99^2=(10^n-1)^2=10^{2n}-2\cdot10^n+1$. The other two families are obtained by dividing by $3^2$ and multiplying by $2^2$. Taking the second family, we have $$$$ $$\begin{align} {\underbrace{66...66}_n}^2 & ={\underbrace{44...44}_{n-1}}3{\underbrace{55...55}_{n-1}}6 \\ {\underbrace{66...6}_{n-1}7}^2 & = ({\underbrace{66...66}_n}+1)^2 \\ & = {\underbrace{44...44}_{n-1}}3{\underbrace{55...55}_{n-1}}6 + 2\cdot{\underbrace{66...66}_n}+1 \\ & = {\underbrace{44...44}_{n-1}}3{\underbrace{55...55}_{n-1}}6 + 1{\underbrace{33...33}_{n}} \\ & = {\underbrace{44...444}_{n}}{\underbrace{88...88}_{n-1}}9 \end{align}$$

So we have $${\underbrace{66...66}_n}^2={\underbrace{44...44}_{n-1}}3{\underbrace{55...55}_{n-1}}6 < \underbrace{44...44}_{2n-1}1 < {\underbrace{44...444}_{n}}{\underbrace{88...88}_{n-1}}9 = {\underbrace{66...6}_{n-1}7}^2$$

Answer 3

The answer of @Aig correctly chose to look at the given number in mod 11, but I would like to show why mod 11 was specifically chosen.

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Solution

A number of form $aaa \dots aa1$ with $2n$ digits ($2n-1$ a's) can be written as

$$\underbrace{aaa \dots aa1}_{2n} = \sum_{k=1}^{2n-1} a(10^k) + 1$$

Which can be written as

$$= a(10^{2n-1}) + a(10^{2n-2}) + \dots + a(10) + 1$$

Adding and subtracting $(a-1)$

$$= \left[ a(10^{2n-1}) + a(10^{2n-2}) + \dots + a(10) + a \right] - (a-1)$$

$$= a\left[(10^{2n-2})(10 + 1) + (10^{2n-4})(10 + 1) + \dots + (1)(10 + 1)\right] - (a-1)$$

$$= 11a\left[10^{2n-2} + 10^{2n-4} + \dots + 1 \right] - (a-1)$$

If we look at it in mod 11, then

$$\underbrace{aaa \dots aa1}_{2n} \equiv -(a-1) \pmod{11}$$ $$-(a-1) \equiv 12 - a \pmod{11}$$

Using the table in the answer of @Aig, $\underbrace{aaa \dots aa1}_{2n}$ is not a perfect square when

$$12-a \not\equiv \{0,1,4,9,5,3\} \pmod{11}$$ $$\implies a \not\equiv \{12,11,8,3,7,9 \} \pmod{11}$$ $$\implies a \not\equiv \{1,0,8,3,7,9 \} \pmod{11}$$

As $a$ is a digit, we get

$$a = \{ 2,4,5,6 \}$$

Answer

Hence all numbers of form $aaa \dots aa1$ where $a = {2,4,5,6}$ cannot be perfect squares

Method

Modular Arithmetic is often a handy tool when solving a wide variety of questions in Number Theory. I am afraid I do not possess the required knowledge to answer your other 2 side questions.