Finding perfect squares

#NumberTheory#Squares

Problem Statement

Let $\lbrace a_{n}\rbrace_{n\ge1}$ be a sequence of natural numbers defined by

$$ a_{1}=1, \ a_{2} = 3, \ \text{and for every integer} \ n \ge 3, $$

$$ a_{n} = (n+1) a_{n-1} - n a_{n-2}. $$

Determine all positive integers $n$ for which $a_{n}$ is a perfect square.

The given recurrence can be rewritten as

$$ (a_{n}-a_{n-1}) = n(a_{n-1}-a_{n-2}), \ \text{for} \ n \ge 3. $$

Define $ a_{n}-a_{n-1} = b_{n}$, so we have $b_{2}=2$. The given recurrence becomes

$$ b_{n} = n b_{n-1} \implies \prod_{r=3}^{n} \frac{b_{r}}{b_{r-1}} = \prod_{r=3}^{n} r $$

$$ \implies \frac{b_{n}}{b_{2}} = 3 \cdot 4 \cdots n \implies b_{n}=n! $$

Returning to the original recurrence, we have the following.

$$ a_{r} - a_{r-1} = r! $$

$$ \sum_{r=2}^{n} (a_{r} - a_{r-1}) = \sum_{r=2}^{n} r! $$

$$ \implies a_{n} - a_{1} = \sum_{r=2}^{n} r! $$

$$ \implies a_{n} = \sum_{r=1}^{n} r! $$

Clearly $a_{1},a_{3}$ are perfect squares, whereas $a_{2}$ and $a_{4}$ are not. For $n \ge 5$, we have $a_{n} \equiv 3 (\mod{10}) $ and therefore can never be a perfect square. So, the only perfect squares are $a_{1}$ and $a_{3}$.