Grigorios Kostakos wrote:

Prove that every term of the sequence \begin{align*} &49\,,\quad 4489\,,\quad 444889\,,\quad 44448889\,,\\ &4444488889\,, \quad 444444888889\,,\ldots \end{align*} is a perfect square.

Hello Grigoris (Αρχιμήδης 6 greets you),

the numbers appearing in the sequence above are of the form:

$$\begin{aligned}

a_k &=4\left ( 10^{2k-1}+10^{2k-2}+\cdots +10^k \right )+ 8\left ( 10^{k-1}+\cdots +10 \right )+9 \\

&=4\left ( 10^k \cdot \frac{10^k-1}{10-1} \right )+ 8 \left ( 10 \cdot \frac{10^{k-1}-1}{10-1}\right )+9 \\

&= \frac{4}{9}\left ( 10^{2k}-10^k \right )+\frac{8}{9} \left ( 10^k-10 \right )+9 \\

&= \left ( \frac{2\cdot 10^k +1}{3} \right )^2

\end{aligned}$$

that are indeed perfect squares. Also note that each term is an integer since $$\displaystyle 10^k \equiv 1 \bmod 3, \; \;\; 2 \cdot 10^k +1 \equiv 0 \bmod 3 $$

Hence \( \dfrac{2\cdot 10^k+1}{3} \in \mathbb{Z} \).