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Work in the prime field ${\mathbb{F}}_p$ and set $r=p-1$. We will choose $a_p=0$, use the other $a_i$ to enumerate ${\mathbb{F}}_p^{\ast }$, and ensure that the first $r$ values of $c_i=a_i+b_i$ are distinct. The zero-sum condition will then force the last value to be the missing field element.

Consider the polynomialIt has total degree of a polynomial $r(r-1)$. Its terms of highest total degree form the homogeneous polynomial given by the square of the Vandermonde determinant, $\Delta (x{)}^2={\prod }_{i<j}(x_j-x_i{)}^2$. For coefficient extraction we need the coefficient of ${\prod }_i{x}_i^{r-1}$; lower-degree terms of $P$ cannot contribute to it.

Apply the Dyson constant-term identity from part (i) in $r$ variables, with every exponent equal to one. Pairing its factors for $i,j$ givesThe constant term is $r!$. Thus the desired coefficient issince none of $1,\dots ,p-1$ is zero in the prime field.

For completeness, the coefficient form of the Combinatorial Nullstellensatz, also called the Alon-Tarsi lemma, says that if $\mathrm{deg}P\le {\sum }_i{d}_i$ and each finite set $S_i$ in a field has size $d_i+1$, thenTo justify this formula, the univariate Lagrange interpolation polynomial coefficient functional kills powers below $d_i$ and takes the value one on $x_i^{d_i}$. Apply the product of these functionals to each monomial of $P$. Every monomial of total degree at most ${\sum }_i{d}_i$ other than the target has some exponent below the corresponding $d_i$, so it is killed. The target survives with coefficient one. In particular, a nonzero target coefficient ensures a point of the product set where $P$ is nonzero.

Use $S_i={\mathbb{F}}_p^{\ast }$ and $d_i=r-1$ for every $i$. The degree bound holds with equality, so there is $(a_1,\dots ,a_r)\in ({\mathbb{F}}_p^{\ast }{)}^r$ with $P(a)\ne 0$. Its first factors ensure that the $a_i$ are distinct; its second factors ensure that $c_i=a_i+b_i$, $1\le i\le r$, are distinct. With $a_p=0$, the $a_i$ enumerate the whole prime field.

Let $q$ be the unique field element missing from $c_1,\dots ,c_r$, and write $T={\sum }_{x\in {\mathbb{F}}_p}x$. Since ${\sum }_i{b}_i=0$,The same sum is $T-q$, hence $q=b_p$. Taking $c_p=b_p=a_p+b_p$ completes the enumeration. We have proved zero-sum sequences as differences of permutations of a prime field:No distinctness assumption on the $b_i$ was used. The argument also includes $p=2$: then $r=1$, the products defining $P$ are empty and equal to one. Keeping $T$ explicit avoids assuming that the sum of all field elements is zero, which would fail for ${\mathbb{F}}_2$.

The coefficient form of the Combinatorial Nullstellensatz is the following. Let $f\in F[x_1,\dots ,x_n]$ have total degree at most $d_1+\cdots +d_n$, and supposeFor arbitrary subsets $S_i\subseteq F$ with $|S_i|=d_i+1$, there is an $x\in S_1\times \cdots \times S_n$ such that $f(x)\ne 0$.

For the proof, defineSuccessive Lagrange interpolation in the variables gives the Alon-Tarsi lemmaEvery denominator is nonzero because the elements of $S_i$ are distinct. If $f$ vanished throughout the product grid, the right side and hence the assumed nonzero coefficient would vanish. This contradiction proves the theorem.

The Alon-Tarsi lemma says that if $\mathrm{deg}f\le d_1+\cdots +d_n$ and $A_i$ consists of $d_i+1$ distinct field elements, thenThis follows by applying univariate Lagrange interpolation successively in each variable.

If the displayed coefficient is nonzero, at least one summand has $f(a_1,\dots ,a_n)\ne 0$. This is the coefficient form of the Combinatorial Nullstellensatz.