Functions which satisfy $f(n)+2f(f(n))=3n+5$ I need to find all functions that satisfy such relation:
$$f(n)+2f(f(n))=3n+5$$
And we knew that the function is defined on the set of positive integers.
So I have calculated that $f(1)=2$, $f(2)=3$, $f(3)=4$.
Observing from this pattern, I could use the arithmetic formula to find one function that satisfying this relation; $a_n=a+(n-1)d$ Substituting $a=1, d=1$:$$a_n=1+n.$$
May I know is there any more functions that satisfy such relation?  Or there is only one function, which is $n+1$ satisfying the relation.
 A: Induction! Suppose $f(n)=n+1$ Then, $$f(n)+2f(f(n))=3n+5$$
is equivalent to $$n+1+2f(n+1)=3n+5$$
so we get $$2f(n+1)=2(n+2)$$
So $f(n+1)=n+2$, which is what we wanted.
Moreover, as you calculated $f(1)=2$, our induction is complete. (note that this is a very important step)
A: Here is an interesting solution given by user littletush in https://artofproblemsolving.com/community/c6h446713p2514093.
If we let $g(n)=f(n)-n-1$ and substitute it to the functional equation (this is motivated by observing that $f(n)=n+1$ is at least one of solutions), we get
$$3g(n)=-2g(g(n)+n+1).$$
However then we see $2\mid g(n)$ for all $n$, which in turn means $2 \mid g(g(n)+n+1)$, which in turn means $4 \mid g(n)$, and so on, we can easily prove that generic $2^k \mid g(n)$ for all $k, n$. Only integer that can satisfy this property is $0$ (remember values of $g$ can be also non-positive integers, as opposed to values of $f$), thus we have $g(n)=0$ and so $f(n)=n+1$ follows.
A: Let $f(n)=an+b$, then $$f(n)+2f(f(n)=3n+5 \implies an+b+2(a(an+b)+b)=3n+5$$
$$\implies 2a^2+a=3, a+2ab+2b=5$$ $$ \implies a=1, b=1.$$ So $f(n)=n+1.$
The other solution for $a=-3/2$ gives $0=5$ which is invalid.
A: Assuming a linear function $f(n)=an+b$,
$$an+b+2a^2n+2ab+2b=3n+5$$ and $$\begin{cases}a+2a^2=3,\\(3+2a)b=5\end{cases}$$
The only solution is $a=b=1$.
Of course this says nothing about possible nonlinear solutions. A polynomial cannot do.

As explained by @Vlad, we can prove by induction that $f(n)=n+1$.
Indeed,
$$f(n)=n+1\implies n+1+2f(n+1)=3n+5\implies f(n+1)=n+2$$ and $f(1)=2$ is known. So this solution is guaranteed.
