# Show that for each positive integer $n$, $a_n$ is a positive integer, $a_{n+1} = (a_n+1)a+(a+1)a_n+2\sqrt{a(a+1)a_n(a_n+1)} \quad (n = 1,2,\ldots).$

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

Firstly, why do they say $a_0 = 0$ if we want to show that for each positive integer $n$, $a_n$ is a positive integer?

The solution in my book did the following. Can someone explain how they get the results "by induction"? In particular, how do they get $\sqrt{a_{n+1}}-\sqrt{a_n} = \left(\sqrt{a+1}-\sqrt{a}\right)^n$ using the results from part 1 and induction?

• @πr8 How do we prove the result by induction? The fact that the sequence doesn't include $n = 0$ I think makes it harder. – user19405892 Jul 9 '16 at 3:10
• How do we prove that we have $b_n = \left(\sqrt{a+1}-\sqrt{a}\right)^n$? – user19405892 Jul 9 '16 at 3:18
• @πr8 How do we solve that? Also, the solution said to use induction and so do we use it? – user19405892 Jul 9 '16 at 3:24

If we have $\sqrt{a_{n+1}+1}-\sqrt{a_{n+1}}=(\sqrt{a+1}-\sqrt{a})(\sqrt{a_{n}+1}-\sqrt{a_{n}}$ for all $n\geq0$, then \begin{align} \sqrt{a_{n}+1}-\sqrt{a_{n}}=\frac{\sqrt{a_{n}+1}-\sqrt{a_{n}}}{\sqrt{a_{n-1}+1}-\sqrt{a_{n-1}}}\frac{\sqrt{a_{n-1}+1}-\sqrt{a_{n-1}}}{\sqrt{a_{n-2}+1}-\sqrt{a_{n-2}}}\cdots\frac{\sqrt{a_{1}+1}-\sqrt{a_{1}}}{\sqrt{a_{0}+1}-\sqrt{a_{0}}}=(\sqrt{a+1}-\sqrt{a})^{n} \end{align}
• Your proof is exact, but couldn't we say, in a simpler way, that one switches from the recurrent definition of a geometrical sequence $b_{n+1}=Kb_n$ to its explicit form : $b_n=K^n b_0$ (with $b_n:=\sqrt{a_{n}+1}-\sqrt{a_{n}}$) ? – Jean Marie Jul 9 '16 at 3:02