Finding $f(2020)$ given that $f(n+m) + f(n-m) = f(3n)$

Question

Assume we have a function $f$ such that $\mathbb Z^{0+} \rightarrow R,$ and $f(n+m) + f(n-m) = f(3n).$ If $n,m \in \mathbb Z^{0+},$ find $f(2020).$

My immediate thought process for this was to substitute small values of $n$ and $m$ in order to try and find $f(2020),$ but I realized that $\frac{2020}{3}$ wasn't an integer and therefore I couldn't just substitute in $n = \frac{2020}{3}.$ Are there any other ways to begin on this problem?

Hint. What is $f(0)$? Then compare $m = 1$ with $m = 0$, for any $n$. What can you conclude? — Brian Tung, Aug 20 '20 at 01:27
Well, if we let $n = 0,$ than $f(0)$ should be equal to $f(m) + f(-m).$ — questionasker, Aug 20 '20 at 01:34
$f$ isn't defined for negative arguments, so in order to make sense of your last statement, $m$ must be...? — Brian Tung, Aug 20 '20 at 03:53

score 2 · Accepted Answer · answered Aug 20 '20 at 01:27

For any $n \ge 1$, we have $$ f(n+1) + f(n-1) = f(3n) = f(n) + f(n) $$ and therefore $f(n+1) = 2f(n) - f(n-1)$. This is a two-term linear recurrence to which all solutions have the form $f(n) = an + b$. Substituting this into the functional equation gives us $$ 2an + 2b = 3an + b \implies an = b $$ which can only hold for all $n$ if $a=b=0$. Therefore $f(2020)=0$ as well.

Finding $f(2020)$ given that $f(n+m) + f(n-m) = f(3n)$

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