The following series seems to become an arithmetic series $$a_{n+2} = a_n^2-n\cdot a_{n+1}$$ If $$a_1=3, a_2=4$$ $$(a_3=5, a_4=6, ..., a_n = n+2)$$
Can $a_n=n+2$ be derived from the original recursion equation?
And are there any other initial values that will also lead to an arithmetic series?
On
If the terms form an arithmetic sequence,
$$a_n=\alpha n+\beta.$$
Then by the recurrence,
$$\alpha(n+2)+\beta=(\alpha n+\beta)^2-n(\alpha(n+1)+\beta)=(\alpha^2-\alpha)n^2+(2\alpha\beta-\alpha-\beta)n+\beta^2.$$
The term in $n^2$ can only vanish when $\alpha=0$ or $\alpha=1$, which give $$\beta=-\beta n+\beta^2\implies \beta=0,$$
or $$n+2+\beta=(\beta-1)n+\beta^2\implies \beta=2.$$
There are no other possibilities.
Of course.
If $a_n=n+2$ and $a_{n+1}=(n+1)+2$ then $$a_{n+2}=(n+2)^2-n(n +3)=n+4. $$
For the generalizing question, assume $a_n=bn+c$ for all $n$. Then $$b(n+2)+c = (bn+c)^2-n(b(n+1)+c), $$ i.e., $$bn+(2b+c)=(2bc-b-c)n+c^2$$ for all $n$, which implies $$ b=2bc-b-c,\qquad 2b+c=c^2,$$ hence $2bc=2b+c=c^2 $. One trivial solution is $b=c=0$ (i.e., $a_n=0$ for all $n$). If $c\ne 0$, we are led to $c=2b$ and $c=2$ (i.e., $a_n=n+2$for all $n$).