Sequence : $f(f(n))+f(n+1)=3n$

Question

Does there exist a function $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $f(f(n))+f(n+1)=3n$, $\forall n \in \mathbb{Z}^+$ ?

My attempt :

Substitute $n=1$, $f(f(1))+f(2)=3$ so $f(2) \in \{1$ or $2\}$

If $f(2)=1, f(f(1))=2$

Substitute $n=2$, $f(f(2))+f(3)=6$ so $f(1)+f(3)=6$

Substitute $n=3$, $f(f(3))+f(4)=9$

If $f(2)=2$, $f(f(1))=1$ so $f(1)\not= 2$

Answer 1

I claim that no such functions exist.

First, suppose $f(2)=2$. We get $$f(f(2))+f(3)=6\Longrightarrow f(3)=4.$$ We get a contradiction, since $$2f(4)=f(f(3))+f(4)=9.$$

Hence $f(2)=1$. As you said, $f(f(1))=2$, and in particular $f(1)\neq 2$. Let's go through all the cases of $f(1)$ (since $f(1)+f(3)=6$, there are only 5 cases):

• $f(1)=1$; then $2=f(f(1))=f(1)=1$, a contradiction.
• $f(1)=2$ was already ruled out.
• $f(1)=3$; then $f(3)=3$. But also $f(3)=f(f(1))=2$, a contradiction.
• $f(1)=4$; then $f(4)=f(f(1))=2$ and $f(3)=2$. We get a contradiction, since $9=f(f(3))+f(4)=f(2)+f(4)=4$.
• $f(1)=5$; then $f(5)=f(f(1))=2$ and $f(3)=1$. Hence $$9=f(f(3))+f(4)=f(1)+f(4)=5+f(4),$$ implying $f(4)=4$. This does not work with $f(f(4))+f(5)=12$.

As a conclusion, there are no such functions.

Answer 2

We will show that there is no such function.

Since $f(2)+ff(1)=3$, we have two case:

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$f(2)=2$:

$ff(2)+f(3)=6\implies f(3)=4$

$ff(3)+f(4)=9\implies 2f(4)=9$.

The last one is impossible hence $f(2)=1$.

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$f(2)=1:$

$$ ff(2)+f(3)=6\implies f(3)=6-f(1)\\ ff(3)+f(4)=9\implies f(6-f(1))+f(4)=9. $$ First of all $f(1)\neq 1,2$.

If $f(1)=5$ then : $$ ff(1)=2\implies f(5)=2\\ f(3)=1\implies f(4)=4\\ ff(4)+f(5)=4+f(5)=4+2\neq 3\times 4. $$ So $f(1)\neq 5$.

If $f(1)=3$: $$ ff(1)=f(3)=2\\ f(3)=6-f(1)=3 $$ So again $f(1)\neq 3$.

If $f(1)=4$ then $f(3)=2$: $$ f(f(1))=f(4)=2\\ ff(3)+f(4)=9\implies 1+f(4)=9. $$ So $f(1)\neq 4$.

Therefore there is no function satisfying this condition.