If \(f(x_1,\cdots,x_n)\) has degree \(d\) less than the number of variables. Then, the number of solutions to $$ \begin{align*} f(x_1,\cdots,x_n) \equiv 0 \pmod{m} \end{align*} $$ is divisible by \(p\).

One application of this is: Suppose that the constant term is \(0\), then \(x_1 = x_2 = \cdots = x_n = 0\) is a solution. But since the number of solutions is divisible by \(p\), then this implies that there must be another solution. In fact, this means that there are at least \(p-1\) solutions.

Before proving the theorem, we need the following useful lemma

If \(p\) is prime and \(i < p-1\), then $$ \begin{align*} 0^i + 1^i + 2^i + \cdots + (p-1)^i \equiv 0 \pmod{p} \end{align*} $$

Let’s first try out small values of \(p\) to see if this makes sense. Take \(p = 3\). Then

\(x/i\)\(0\)\(1\)\(2\)
\(0^i\)\(1\)\(0\)\(0\)
\(1^i\)\(1\)\(1\)\(1\)
\(2^i\)\(1\)\(2\)\(4\)
sum\(3\)\(3\)\(5\)

Notice now that for the values less than \(p\), we get \(0 \pmod{p}\) while for \(i \geq p-1\) like \(i=2\), we get \(5\) and that doesn’t vanish modulo \(3\).

Lemma Proof

Let \(S = 0^i + 1^i + 2^i + \cdots + (p-1)^i\). We want to show that \(S \equiv 0 \pmod{p}\) when \(i < p - 1\). Now, pick \(a\) such that

$$ \begin{align*} a \not\equiv 0 \pmod{p}, \quad \text{ and } \quad a^i \not\equiv 1 \pmod{p} \end{align*} $$

Let’s check why we can do this. Recall in the previous lecture we said that if \(d\) divides \(p-1\), then \(x^d - 1 \equiv 0 \pmod{p}\) has exactly \(d\) roots. Here, we have \(d = i\) so we have exactly \(i\) roots. But by assumption \(i < p - 1\) and we have \(p-1\) values that are not zero: \(1,2,\cdots,(p-1)\). Therefore, one of the values \(1,2,\cdots,(p-1)\) is not a root. Thus \(a^i - 1 \not\equiv 0 \pmod{p}\) or as written above \(a^i \not\equiv 1 \pmod{p}\).

Take the values \(1,2,\cdots,(p-1)\) and multiply them by \(a\). Multiply each value by \(a\)

$$ \begin{align*} 0, 1a, 2a, \cdots, a(p-1) \pmod{p} \end{align*} $$

These values remain the same but only permuted since \(a\) has an inverse modulo \(p\). Now, let \(S\) be the sum of the ith power of each one of these values. So

$$ \begin{align*} S = 0^i + (1)^i + (2)^i + \cdots + (p-1)^i \pmod{p} \end{align*} $$

Then \(S\) should be the same as the sum

$$ \begin{align*} &= 0^i + (1a)^i + (2a)^i + \cdots + (a(p-1))^i \pmod{p} \\ &= a^i [0^i + (1)^i + (2)^i + \cdots + (p-1)^i] \pmod{p} \\ &= a^i S \end{align*} $$

So this means that

$$ \begin{align*} S &\equiv a^i S \pmod{p} \\ a^iS - S &\equiv 0 \pmod{p} \\ S(a^i - 1) &\equiv 0 \pmod{p} \end{align*} $$

Now recall that we choose \(a\) such that \(a^i \not\equiv 1 \pmod{p}\). Furthermore, since \(p\) is prime, then we have no zero divisors. This means that we must have

$$ \begin{align*} S &\equiv 0 \pmod{p} \end{align*} $$

as we wanted to show. \(\blacksquare\)

Theorem Proof

So \(f(x_1,\cdots,x_n)\) be a polynomial in \(n\) variables with \(\deg f < n\). Observe now that

$$ \begin{align*} 1 - f(x_1,\cdots,x_n)^{p-1} \equiv \begin{cases} 1 & \text{ if } f(x_1,\cdots,x_n) \equiv 0 \pmod{p}, \\ 0 & \text{if } f(x_1,\cdots,x_n) \not\equiv 0 \pmod{p}. \end{cases} \end{align*} $$
This is true by Fermat's Little Theorem. To see this, let \(a = f(x_1,\cdots,x_n)\) Then, if \(a \equiv 0 \pmod{p}\), then $$ \begin{align*} 1 - a^{p-1} \equiv 1 - 0^{p-1} \equiv 1 \pmod{p} \end{align*} $$ and if \(a \not\equiv 0 \pmod{p}\), then by Fermat, we we must have $$ \begin{align*} a^{p-1} \equiv 1 \pmod{p} \\ 1 - a^{p-1} \equiv 0 \pmod{p} \\ \end{align*} $$

So this says, the expression \(1 - f(x_1,\cdots,x_n)^{p-1}\) is only congruent to \(0\) when \(x_i\) is not a root. Therefore

$$ \begin{align*} \sum_{x_1,x_2,\cdots,x_n} 1 - f(x_1,\cdots,x_n)^{p-1} \equiv \end{align*} $$

References