How prove this sequence $a_{n}$ is $a_{n}=2n+1$ without mathematical induction? 
Let  sequence  $a_{1}=3$ and $$a_{n+1}=a^2_{n}-2na_{n}+2,n\ge 1$$
Find the $a_{n}$

My idea: 
since
$$a_{2}=9-6+2=5,a_{3}=a^2_{2}-4a_{2}+2=25-20+2=7,a_{4}=a^2_{3}-6a_{3}+2=49-42+2=9$$
so I inductive
$$a_{n}=2n+1$$
and it is easy use Mathematical induction to prove it
since
$$a_{n+1}=(2n+1)^2-2n\cdot (2n+1)+2=2(n+1)+1$$
My Question,This sequence have without mathematical induction methods?
such as this two  post. At first I thought to can't find other ways, but finally surprised someone gives three different methods:
How find this $n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}$
,How find $a_{n}$ if the sequence $a_{n}=2a_{n-1}+(2n-1)^2a_{n-2},n\ge 1$

Let $b_n = a_n - 2n$. Observe that $b_1 = 1$. We can substitute this into the given recurrence relation and get $$b_{n+1} + 2n = (b_n + 2n)\cdot b_n$$ In particular, this formula tells us: if for any $n$ we have $b_n = 1$, then $b_{n+1} = b_n$. Since $b_1 = 1$, it follows the sequence $\{b_i\}_{i=1}^{\infty}$ is constant. Hence $a_n = 2n + b_n = 2n + 1$.
