### 1986 IMO SL #7

Let $a$ be a postive integer and let ${a_n}$ be defined by $a_0=0$ and \[a_{n+1}=(a_n+1)a+(a+1)a_n+2\sqrt{a(a+1)a_n(a_n+1)}\]Show that for each positive integer $n,a_n$ is a positive integer.

Clearly $a_n$ is increasing and positive. Note $$\sqrt{a_{n+1}}=\sqrt{(a+1)a_n}+\sqrt{a(a_n+1)}.$$ Rearranging and solving a quadratic, we see that $$\sqrt{a_{n-1}}=\sqrt{(a+1)a_n}-\sqrt{a(a_n+1)}.$$ Then, $a_{n-1}a_{n+1}=(a_n-a)^2$ and $a_{n+1}+a_{n-1}=2(2aa_n+a+a_n)$. Hence if $a_{n-1}$ and $a_n$ are integers, so is $a_{n+1}$. But $a_0=0$ and $a_1=a$, so by induction we conclude that all $a_n$ are positive integers, as desired. $\square$

oh this was a really nice way to finish it

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