Which is the two-variable polynomial $p$ such that

$$
p(x+y, x y)=\sum_{k=0}^{20} x^{20-k} y^{k} ?
$$
Solution. Let us use the notation

$$
S_{k}=x^{k}-y^{k}
$$

for all positive integers $k$; let $x+y=a$ and $xy=b$. According to the known identity,

$$
\sum_{k=0}^{20} x^{20-k} y^{k}=\frac{S_{21}}{x-y}=\frac{S_{21}}{S_{1}}
$$

If $n>1$, then

$$
S_{n+1}=x^{n+1}-y^{n+1}=\left(x^{n}-y^{n}\right)(x+y)-\left(x^{n} y-x y^{n}\right)=a S_{n}-b S_{n-1}
$$

Applying this for $n=1,2, \ldots, 6$:

$$
\begin{aligned}
& S_{1}=x-y \\
& S_{2}=a(x-y)=a S_{1} \\
& S_{3}=a^{2}(x-y)-b(x-y)=\left(a^{2}-b\right) S_{1} \\
& S_{4}=a\left(a^{2}-b\right) S_{1}-b a S_{1}=\left(a^{3}-2 a b\right) S_{1} \\
& S_{5}=a\left(a^{3}-2 a b\right) S_{1}-b\left(a^{2}-b\right) S_{1}=\left(a^{4}-3 a^{2} b+b^{2}\right) S_{1} \\
& S_{6}=a\left(a^{4}-3 a^{2} b+b^{2}\right) S_{1}-b\left(a^{3}-2 a b\right) S_{1}=\left(a^{5}-4 a^{3} b+3 a b^{2}\right) S_{1} \\
& S_{7}=a\left(a^{5}-4 a^{3} b+3 a b^{2}\right) S_{1}-b\left(a^{4}-3 a^{2} b+b^{2}\right) S_{1}=\left(a^{6}-5 a^{4} b+6 a^{2} b^{2}-b^{3}\right) S_{1}
\end{aligned}
$$

Finally,

$$
S_{21}=x^{21}-y^{21}=\left(x^{7}\right)^{3}-\left(y^{7}\right)^{3}=\left(x^{7}-y^{7}\right)\left(\left(x^{7}\right)^{2}+x^{7} y^{7}+\left(y^{7}\right)^{2}\right)=S_{7}\left(S_{7}^{2}+3 b^{7}\right)
$$

Thus,

$$
\begin{aligned}
\frac{S_{21}}{x-y} & =\left(a^{6}-5 a^{4} b+6 a^{2} b^{2}-b^{3}\right)^{3} S_{1}^{2}+3 b^{7}\left(a^{6}-5 a^{4} b+6 a^{2} b^{2}-b^{3}\right)= \\
& =\left(a^{6}-5 a^{4} b+6 a^{2} b^{2}-b^{3}\right)^{3}\left(a^{2}-4 b\right)+3 b^{7}\left(a^{6}-5 a^{4} b+6 a^{2} b^{2}-b^{3}\right)
\end{aligned}
$$

Therefore, the desired polynomial is

$$
\begin{aligned}
p(z, t)= & \left(z^{6}-5 z^{4} t+6 z^{2} t^{2}-t^{3}\right)^{3}\left(z^{2}-4 t\right)+3 t^{7}\left(z^{6}-5 z^{4} t+6 z^{2} t^{2}-t^{3}\right)= \\
= & z^{20}-19 z^{18} t+153 z^{16} t^{2}-680 z^{14} t^{3}+1820 z^{12} t^{4}-3003 z^{10} t^{5}+3003 z^{8} t^{6}- \\
& -1716 z^{6} t^{7}+495 z^{4} t^{8}-55 z^{2} t^{9}+t^{10}
\end{aligned}
$$

Remark. Generally, for polynomials $p_{n}=p_{n}(z, t)$ satisfying the condition

$$
p_{n}(x+y, x y)=\sum_{k=0}^{n} x^{n-k} y^{k}
$$

the recursion

$$
p_{n+1}(z, t)=z p_{n}(z, t)-t p_{n-1}(z, t)
$$

analogous to the recursion $S_{n+1}=a S_{n}-b S_{n-1}$ used in the solution, can be shown by induction on $n$ that

$$
p_{n}(z, t)=\sum_{i=0}^{[n / 2]}(-1)^{i}\binom{n-i}{i} z^{n-2 i} t^{i}
$$