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Given a sequence $(a_n)$, with $a_1 = 4$ and $a_{n+1} = a_n^2-2$ for all $n \in \mathbb{N}$, prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer.

We must have that \begin{align*}(a_n-1)+a_n &> a_n+1\\a_n+(a_n+1) &> a_n-1\\(a_n-1)+(a_n+1) &> a_n,\end{align*} which gives us equivalently that $a_n > 2$. Now using Heron's Formula we get \begin{align*}S = \sqrt{\dfrac{3a_n}{2}\left(\dfrac{3a_n}{2}-(a_n-1)\right)\left(\dfrac{3a_n}{2}-a_n\right)\left(\dfrac{3a_n}{2}-(a_n+1)\right)} &= \sqrt{\dfrac{3a_n}{2} \cdot \dfrac{a_n+2}{2} \cdot \dfrac{a_n}{2} \cdot \dfrac{a_n-2}{2}}\\&=\dfrac{1}{4}\sqrt{3a_n^2(a_n-2)(a_n+2)}.\end{align*} How do we continue from here?

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You have proven that $$S_n = \frac{1}{4}\sqrt{3a_n^2(a_n^2-4)}$$

Suppose $S_n$ is an integer.

$$S_n^2 = \frac{3}{16}a_n^2(a_n^2-4)$$

We have

\begin{align} S_{n+1}^2 &= \frac{3}{16}a_{n+1}^2(a_{n+1}^2-4)\\ & = \frac{3}{16}(a_{n+1}^2-4) a_{n+1}^2\\ & = \frac{3}{16}((a_{n}^2-2)^2-4) a_{n+1}^2\\ &= \frac{3}{16}(a_n^4-4a_n^2) a_{n+1}^2\\ &= S_n^2a_{n+1}^2\\ \end{align}

That is we have $$S_{n+1}=S_na_{n+1}$$

Checking $a_n$ are integers are simple. Also, for base case, $S_1=6$ which is an integer.

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We see that $3a_n^2(a_n^2-4)$ must be a perfect square divisible by $4$. We notice that $a_n \equiv 2 \pmod{4}$ and so $a_n^2$ is a multiple of $4$, similarly with $a_n-2$ and $a_n+2$. Thus we just need $3a_n^2(a_n^2-4)$ to be a perfect square. To prove this, we prove the following: $$a_n^2(a_n^2-4) = 12(a_1 \cdots a_n)^2 \quad \text{for } n \geq 1.$$ We prove this by induction on $n$. For $n = 1$ the result holds trivially since $a_1^2(a_1^2-4) = 4^2(4^2-4) = 12(a_1)^2 = 12(4)^2$. Now suppose the result holds for some $n$. Then noting that $$a_n^2(a_n^2-4) = a_{n+1}^2-4,$$ we see that $a_{n+1}^2(a_{n+1}^2-4) = 12(a_1 \cdots a_{n+1})^2$, completing the induction.

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