Functions which satisfy $f(n)+2f(f(n))=3n+5$

Question

I need to find all functions $f:\mathbb N^*\to\mathbb N^*$ that satisfy the relation: $$f(n)+2f(f(n))=3n+5.$$

Here $\mathbb N^*$ denotes the set of positive integers.
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 if there are other functions that satisfy such relation? Or there is only one function, which is $n+1$ satisfying the relation.

Answer 1

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)

Answer 2

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.

Answer 3

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.

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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.

Answer 4

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.