The problem was to show that the sequence of numbers
$$49, 4489, 444889, \dots $$
where each number is formed by inserting 48 in the middle of the previous number:
$$49 \to 4\underline{48}9 \to 44\underline{48}89 \to \dots$$
has only perfect squares!
Solution
Any number is of the form $n$ 4s followed by $n-1$ 8s followed by 9.
This can be rewritten as the sum of
$4444\dots44$ (a $2n$ digit number)
$44\dots4$ (an $n$ digit number)
1
This is basically
$$\frac{4(10^{2n}-1)}{9} + \frac{4(10^{n}-1)}{9} + 1 $$
$$= \left(\frac{2\times10^n + 1}{3}\right)^2$$
$$49, 4489, 444889, \dots $$
where each number is formed by inserting 48 in the middle of the previous number:
$$49 \to 4\underline{48}9 \to 44\underline{48}89 \to \dots$$
has only perfect squares!
Solution
Any number is of the form $n$ 4s followed by $n-1$ 8s followed by 9.
This can be rewritten as the sum of
$4444\dots44$ (a $2n$ digit number)
$44\dots4$ (an $n$ digit number)
1
This is basically
$$\frac{4(10^{2n}-1)}{9} + \frac{4(10^{n}-1)}{9} + 1 $$
$$= \left(\frac{2\times10^n + 1}{3}\right)^2$$
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