The problem asks to compute $f(2020)$ knowing that
$f(1) = 1$ and $f(m+n)+f(m-n)=\dfrac 12(f(2n) + f(2m))$ for integers $m,n$ such that $m>n\ge 1$.
My try :
I conjectured $f(n) = kn^2$ where $k$ is a random constant which has to be $1$ due to $f(1) = 1$
I plugged different values in $n$ and $m$ trying to prove this by induction but in vain.
it seem there is something missing.
There is no unique answer but assumptions can be made that give a unique answer. Suppose $f(x)$ can be written in the form
$$f(x)=\sum_{i=0}^\infty a_i x^i$$
where $a_i$ are constants that could be zero. Then
$$0=2f(m+n)+2f(m-n)-f(2m)-f(2m)$$
$$=2\sum_{i=0}^\infty a_i\left(\sum_{j=0}^i\binom{i}{j}n^jm^{i-j}\right)+2\sum_{i=0}^\infty a_i\left(\sum_{j=0}^i\binom{i}{j}(-1)^jn^jm^{i-j}\right)$$
$$-\sum_{i=0}^\infty a_i2^in^i-\sum_{i=0}^\infty a_i 2^i m^i$$
$$=a_0(3-n^2)+a_1(2m-2n^2)+a_2*0+a_3(-4m^3-8n^2+12mn^2)+\cdots$$
It is useful for us to examine this expression at $m=n^3$. Then it becomes
$$0=2\sum_{i=0}^\infty a_i\left(\sum_{j=0}^i\binom{i}{j}n^{3i-2j}\right)+2\sum_{i=0}^\infty a_i\left(\sum_{j=0}^i\binom{i}{j}(-1)^jn^{3i-2j}\right)$$
$$-\sum_{i=0}^\infty a_i2^in^i-\sum_{i=0}^\infty a_i 2^i n^{3i}$$
$$=a_0(3-n^2)+a_1(-2n^2+2n^3)+a_2*0+a_3(-8n^2+12n^5-4n^9)+\cdots$$
Basically, we will show that for $i\geq 3$, the coefficient on $a_i$ is a polynomial of $n$ with degree $n^{3i}$. Now that we know what to look for, this is easy to see. For any $i$, the highest power of $n$ in the coefficient of $a_i$ will be
$$2\binom{i}{0}n^{3i-2*0}+2\binom{i}{0}(-1)^0n^{3i-2*0}-2^in^{3i}=2n^{3i}+2n^{3i}-2^in^{3i}=-(2^i-4)n^{3i}$$
For $i\geq 3$, this will always be nonzero. In order to make the next part clear, denote these polynomials in front of $a_i$ by $p_i$ and define $M_i-1$ to be the maximum coefficient (in terms of absolute value) in the polynomials in the set $\{p_0,p_1,\dots,p_i\}$. Further, define $A_i$ to be $\max\{|a_0|,|a_1|,\dots,|a_i|\}$. Then we know
$$\sum_{j=0}^i a_j p_j\leq \sum_{j=0}^i |a_j p_j|\leq \sum_{j=1}^i A_i M_i n^{3j}+A_iM_in^2=A_i M_i\left[ \frac{n^{3i+3}-n^3}{n^3-1}+n^2\right]$$
But this implies
$$\lim_{n\to\infty}\frac{A_i M_i\left[ \frac{n^{3i+3}-n^3}{n^3-1}+n^2\right]}{p_{i+1}}=\lim_{n\to\infty}A_iM_i \frac{1}{n^3-1}=0$$
We conclude that for all $i\geq 3$, there exists an $N$ such that $n\geq N$ implies
$$\left|\sum_{j=0}^i a_j p_j\right|<|a_{i+1}p_{i+1}|$$
We can manually check that this condition also holds for $i=0$ and $i=2$. Of course, this implies that all the coefficients $a_i$ must be zero (else the expression would not be zero as $n$ goes to infinity). However, there is one exception: $p_2=0$ which means that $a_2$ is in fact a free variable. That is
$$f(x)=a_2x^2$$
From the initial condition, we conclude $f(x)=x^2$. Of course, all of this followed from the assumption that $f(x)$ could be expanded as a (possibly) infinite polynomial. There is not a nice, unique solution if you do not use this assumption. In fact, a whole infinite family of solutions can be constructed quite easily. That is, your solution is uniquely determined by the set $\{f(1),f(2),f(3)\}$. I will now prove this by strong induction:
Set $m=2$ and $n=1$. Then
$$f(4)=f(2m)=-f(2n)+2f(m+n)+2f(m-n)=-f(2)+2f(3)+2f(1)$$
Now, assume we are able to construct $\{f(1),f(2),\dots,f(N)\}$ using a linear combination of $\{f(1),f(2),f(3)\}$ (where $N$ is an even number greater than or equal to $4$). If we can show how to construct $f(N+1)$ and $f(N+2)$, using a linear combination of this set, then we are done. This is easily done by first choosing: $m=\frac{N+2}{2}$ and $n=1$. Then
$$f(N+2)=f(2m)=-f(2n)+2f(m+n)+2f(m-n)$$
$$=-f(2)+2f\left(\frac{N+2}{2}+1\right)+2f\left(\frac{N+2}{2}-1\right)$$
Note that each of these is a linear combination of our set (we assumed this by strong induction) as
$$\frac{N+2}{2}+1=\frac{N}{2}+2\leq N$$
(as $N\geq 4$). For $f(N+1)$, let $m=\frac{N+2}{2}$ and $n=\frac{N}{2}$. Then
$$2f(N+1)=2f(m+n)=f(2m)+f(2n)-2f(m-n)=f(N+2)+f(N)-2f(1)$$
$$f(N+1)=\frac{1}{2}f(N+2)+\frac{1}{2}f(N)-f(1)$$
However, we know all of these can be made as a linear combination of our set. We just showed that $f(N+2)$ can be constructed and we assumed $f(N)$ could be by induction. Thus, every $f(n)$ can be constructed from the set $\{f(1),f(2),f(3)\}$. For example, if
$$\{f(1),f(2),f(3)\}=\{1,4,9\}$$
then $f(x)=x^2$ and $f(2020)=2020^2=4080400$. However, if
$$\{f(1),f(2),f(3)\}=\{1,2,3\}$$
then
$$f(n)=\{1,2,3,6,9,14,17\}\text{ and }f(2020)=1442866$$
If we stick with the set $\{f(1),f(2),f(3)\}$, then it will always be the case that
$$f(2020)=335483 f(1)-211383 f(2)+510050 f(3)$$
Of course, any three initial conditions will produce a unique solution to your functional equation. That is, if you are given
$$\{f(n_1),f(n_2),f(n_3)\}$$
then it is possible to back out $f(x)$ for any value of $x$. This is apparent as
$$f(n_1)=a_1 f(1)+b_1f(2)+c_1 f(3)$$
$$f(n_2)=a_2 f(1)+b_2f(2)+c_2 f(3)$$
$$f(n_2)=a_3 f(1)+b_3f(2)+c_3 f(3)$$
(we just showed this by strong induction). Rearranging and solving for $\{f(1),f(2),f(3)\}$ will give the desired result. Is this system ever unsolvable? Well that depends if
$$\begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}$$
can ever be equal to zero. I do not have a proof, but I believe that it should be possible to show by induction that this is always non-zero. If this were the case, the the functional equation would be completely determined by any three initial values.