6<sqrt41<7 66<sqrt4441<67 666<sqrt444441<667 ...

Theres an really clean solution by one of my teammates: 44….41=44…44-3= 11*404…..04-3. Then consider quadratic residues mod 11.

Tintarn wrote: Can a number of the form $44\dots 41$, with an odd number of decimal digits $4$ followed by a digit $1$, be a perfect square? Assume $\frac{4\cdot 10^{2k}-31}{9}=n^2$. Then we have $\left(2\cdot10^k+3n\right)\left(2\cdot10^k-3n\right)=31$ and thus $2\cdot10^k+3n=31$ and $2\cdot10^k-3n=1$. So $4\cdot10^k=32$ giving us a contradiction.

watereater wrote: Theres an really clean solution by one of my teammates: 44….41=44…44-3= 11*404…..04-3. Then consider quadratic residues mod 8 Could you elaborate? The number is $\equiv 1 \pmod{8}$ which is a quadratic residue, so I don't see how this helps to solve the problem.

Sorry, i meant mod 11

The related question for an even number of digits 4 is also interesting and a bit harder to tackle. Obviously, 1 = 1 x 1 and 441 = 21 x 21, and apart from these, none of the numbers with an even number of decimal digits 4 followed by the digit 1 is a perfect square.

Here is my take on the related question for an even number of digits 4 followed by the digit 1.

Scout66 wrote: Here is my take on the related question for an even number of digits 4 followed by the digit 1.

Good job copying the answers from the offical website

Es gilt: $$44...41 = 1+\sum_{i=1}^{2n+1}4\cdot 10^i=1+4\cdot 10^{2n+1}+\sum_{i=1}^{2n}4\cdot 10^i =1+4 \cdot 10^{2n+1}+4 \cdot \sum_{i=1}^{2n}10^i$$Nun wird die geometrische Reihe benutzt, auf ein Beweis wird hierbei verzichtet. $$1+4 \cdot 10^{2n+1}+4 \cdot 10 \cdot \frac{10^{2n}-1}{10-1}=\frac{4\cdot 10^{2n+2}-31}{9}$$Um zu überprüfen, dass dies eine Quadratzahl ist, wird der Ausdruck mit $(10k+1)^2$ oder $(10k+9)^2$ gleichgesetzt, weil die Quadratzahl auf 1 endet. Daraus folgt: $$(10k+1)^2 = \frac{4\cdot 10^{2n+2}-31}{9}\implies k_{1,2}=-\frac{1}{10}\pm \frac{1}{30}\cdot \sqrt{400\cdot 100^n -31}$$$$(10k+9)^2 = \frac{4\cdot 10^{2n+2}-31}{9}\implies k_{1,2}=-\frac{9}{10}\pm \frac{1}{30}\cdot \sqrt{400\cdot 100^n -31}$$Damit $k$ eine ganze Zahl ist, muss die Diskriminante eine Quadratzahl sein. Wir betrachten dafür zunächst die Reste der Quadratzahlen Modulo 11. Hierfür schauen wir uns die Quadrate der Zahlen von 0 bis 11 an. \newline \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline $x^2$ & $0^2$ & $1^2$ & $2^2$ & $3^2$ & $4^2$ & $5^2$ & $6^2$ & $7^2$ & $8^2$ & $9^2$ & $10^2$ & $11^2$ \\ \hline $x^2 \mod{11}$ & 0 & 1 & 4 & 9 & 5 & 3 & 3 & 5 & 9 & 4 & 1 & 0 \\ \hline \end{tabular} Wir zeigen nun, dass für jede weitere beliebige Quadratzahl $x^2$ gilt: $$ x^2\mod{11} \in \{0,1,3,4,5,9\} $$\begin{align*} x^2 \mod{11} = (x \cdot x) \mod{11} =[(x \mod{11}) \cdot (x \mod{11})] \mod{11} \end{align*}Der Rest $x \mod{11}$ liegt zwischen 0 und 11, wodurch der Rest $x^2 \mod{11}$ gleich dem Rest einer der Quadratzahlen $0^2, ... , 11^2 \mod{11}$ ist. Wir wissen, dass diese Reste in der obigen Menge liegen. Nun zeigen wir, dass die Diskriminante Modulo 11 nicht in der obigen Menge liegt. $$400 \cdot 100^n -31 \mod{11} = [(\underbrace{400 \mod{11}}_{4} \cdot \underbrace{100^n \mod{11}}_{1}) \mod{11} -\underbrace{31 \mod{11}}_{9}]\mod{11}=6~\square$$

Tintarn wrote: Can a number of the form $44\dots 41$, with an odd number of decimal digits $4$ followed by a digit $1$, be a perfect square? This is Oblath problem!

andrejpcuser07 wrote: Good job copying the answers from the offical website I understand that you are concerned about issues of originality or even plagiatarism. Be assured, so am I, and deeply so. Good luck that I had published the solution elsewhere (end of March), before any solutions appeared on the official website (mid April, up to my understanding). Indeed, several authors have published similar solutions in the meantime. I had hoped that the AOPS community would still join me in relishing what I consider a cute approach to the apparently more difficult question for an even number of digits 4.

Let, $a_1 = 41, a_2 = 441, \dots$. Then $a_n = 100a_{n-1} + 331$. Notice that $331$ is prime. Then, it can be proven by induction that $331 \not | a_n$ for $\forall n \in \mathbb{N}$. Suppose, $a_{n-1}$ is not a perfect square. Then $100a_{n-1}$ is not a perfect square and since $331 \not | 100a_{n-1} + 331$, it follows that $a_n$ isn't a perfect square either. Apply induction and we are done.

The answer is no. Let such a number having $2n-1$ decimal digits $4$ be $a_n$. Then clearly $$11 \mid a_{n+1}-a_{n}.$$Therefore, all the $a_i$ are the same $\pmod{11}$. Now observe that $a_1 \equiv 8 \pmod{11}$, so $a_i \equiv 8 \pmod{11} \forall i$. But $8$ is not a quadratic residue $\pmod{11}$. Therefore none of the $a_i$ are perfect squares.

Note we can prove: $(\underbrace{6\cdots 6}_{n})^2=\underbrace{4\cdots 4}_{n-1}3\underbrace{5\cdots 5}_{n-1}6<\underbrace{4\cdots 4}_{2n-1}1<\underbrace{4\cdots 4}_{n}\underbrace{8\cdots 8}_{n-1}9=(\underbrace{6\cdots 6}_{n-1}7)^2$