Bulgarian Mathematical Competitions

Solution:
Considering the polynomials $p(x)=x^{m-1}+x^{m-2}+\cdots+x+1$ and $q(x)=x^{m-1}+x^{m-2}+\cdots+x+a$, $a \neq 1$, shows that the desired minimal number does not exceed $2m-2$ (one has that $f(u, v)=(a-1)\left(u^{m-1}+u^{m-2}+\cdots+u\right)+(1-a)\left(v^{m-1}+v^{m-2}+\cdots+v\right)$). We shall prove by induction on $m$ that the number of the non-zero coefficients is at least $2m-2$.

If $p(x)$ or $q(x)$ contains a monomial which does not appear in the other polynomial, then the non-zero coefficients in $f(u, v)$ are at least $2m$. So we may assume that $p(x)$ and $q(x)$ contain the same monomials. Note also that multiplying some of $p(x)$ and $q(x)$ by a non-zero number does not change the non-zero coefficients of $f(u, v)$.

For $m=2$ one has that $p(x)=a x^{n}+b x^{k}$, $q(x)=c x^{n}+d x^{k}$ and $ad-bc \neq 0$. Then $f(u, v)=(ad-bc) u^{n} v^{k}+(bc-ad) u^{k} v^{n}$ has exactly two non-zero coefficients.

Let $m=3$ and let $p(x)=x^{k}+a x^{n}+b x^{\ell}$, $q(x)=x^{k}+c x^{n}+d x^{\ell}$ and $ad-bc \neq 0$. Then
$$\begin{aligned} f(u, v)= & (ad-bc) u^{\ell} v^{k}+(bc-ad) u^{\ell} v^{n}+(c-a) u^{k} v^{n}+(a-c) u^{n} v^{k} \\ & +(d-b) u^{k} v^{\ell}+(b-d) u^{\ell} v^{k} \end{aligned}$$
The first two coefficients are non-zero. Since the equalities $a=c$ and $b=d$ do not hold simultaneously, then at least two of the last four coefficients are also non-zero.

Let now $m \geq 4$ and let $p(x)=p_{1}(x)+a x^{n}+b x^{k}$, $q(x)=q_{1}(x)+c x^{n}+d x^{k}$, $ad-bc \neq 0$ and any of the polynomials $p_{1}(x)$ and $q_{1}(x)$ has $m-2 \geq 2$ non-zero coefficients. Then $f(u, v)=f_{1}(u, v)+f_{2}(u, v)+f_{3}(u, v)$, where
$$\begin{aligned} f_{1}(u, v)= & p_{1}(u) q_{1}(v)-p_{1}(v) q_{1}(u) \\ f_{2}(u, v)= & \left(a u^{n}+b u^{k}\right) q_{1}(v)+\left(c v^{n}+d v^{k}\right) p_{1}(u) \\ & -\left(a v^{n}+b v^{k}\right) q_{1}(u)-\left(c u^{n}+d u^{k}\right) p_{1}(v) \\ f_{3}(u, v)= & (ad-bc) u^{n} v^{k}+(bc-ad) u^{k} v^{n} \end{aligned}$$
and the different polynomials have no similar monomials. If $p_{1}(x) \neq \alpha q_{1}(x)$, then, by the induction hypothesis, $f_{1}(u, v)$ has at least $2(m-2)-2=2m-6$
non-zero coefficients. Moreover, $f_{2}(u, v)$ has at least two non-zero coefficients and $f_{3}(u, v)$ has two non-zero coefficients.

If $p_{1}(x)=\alpha q_{1}(x)$, $\alpha \neq 0$, then
$f_{2}(u, v)=q_{1}(v)\left[(a-c\alpha) u^{n}+(b-d\alpha) u^{k}\right]+q_{1}(u)\left[(c\alpha-a) v^{n}+(d\alpha-b) v^{k}\right]$.
Since the equalities $a-c\alpha=0$ and $b-d\alpha=0$ do not hold simultaneously, the polynomial $f_{2}(u, v)$ has at least $2m-2$ non-zero coefficients (two times more than those of $q_{1}(x)$). Counting the two non-zero coefficients of $f_{3}(u, v)$, we get the desired result.