Home » Uncategorized » Perfect squares

Perfect squares

Prove that every term of the sequence

    \begin{align*} &49\,,\quad 4489\,,\quad 444889\,,\quad 44448889\,,\\ &4444488889\,, \quad 444444888889\,,\ldots \end{align*}

is a peferct square.

Solution [Αρχιμήδης 6]

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

    \begin{align*} a_k &=4\left ( 10^{2k-1}+10^{2k-2}+\cdots +10^k \right )+ \\ &\quad \quad \quad + 8\left ( 10^{k-1}+\cdots +10 \right )+9 \\ &=4\left ( 10^k \cdot \frac{10^k-1}{10-1} \right )+ \\ & \quad \quad \quad +8 \left ( 10 \cdot \frac{10^{k-1}-1}{10-1}\right )+9 \\ &= \frac{4}{9}\left ( 10^{2k}-10^k \right )+\\ & \quad \quad \quad + \frac{8}{9} \left ( 10^k-10 \right )+9 \\ &= \left ( \frac{2\cdot 10^k +1}{3} \right )^2 \end{align*}

that are indeed perfect squares. Also note that each term is an integer since

10^k \equiv 1 \bmod 3, \; \;\; 2 \cdot 10^k +1 \equiv 0 \bmod 3

Hence \displaystyle \frac{2 \cdot 10^k+1}{3} \in \mathbb{Z}.

Read more

Leave a comment

Donate to Tolaso Network