Subfield: Number Theory


Question: Prove that for every prime $p>100$ and every integer $r$ there exist two integers $a$ and $b$ such that $p$ divides $a^{2}+b^{5}-r$.

Solution:
Throughout the solution, all congruence relations are meant modulo $p$.

Fix $p$, and let $\mathcal{P}=\{0,1, \ldots, p-1\}$ be the set of residue classes modulo $p$. For every $r \in \mathcal{P}$, let $S_{r}=\left\{(a, b) \in \mathcal{P} \times \mathcal{P}: a^{2}+b^{5} \equiv r\right\}$, and let $s_{r}=\left|S_{r}\right|$. Our aim is to prove $s_{r}>0$ for all $r \in \mathcal{P}$.

We will use the well-known fact that for every residue class $r \in \mathcal{P}$ and every positive integer $k$, there are at most $k$ values $x \in \mathcal{P}$ such that $x^{k} \equiv r$.

Lemma. Let $N$ be the number of quadruples $(a, b, c, d) \in \mathcal{P}^{4}$ for which $a^{2}+b^{5} \equiv c^{2}+d^{5}$. Then

$$
N=\sum_{r \in \mathcal{P}} s_{r}^{2}\tag{a}
$$

and

$$
N \leq p\left(p^{2}+4 p-4\right)\tag{b}
$$

Proof. (a) For each residue class $r$ there exist exactly $s_{r}$ pairs $(a, b)$ with $a^{2}+b^{5} \equiv r$ and $s_{r}$ pairs $(c, d)$ with $c^{2}+d^{5} \equiv r$. So there are $s_{r}^{2}$ quadruples with $a^{2}+b^{5} \equiv c^{2}+d^{5} \equiv r$. Taking the sum over all $r \in \mathcal{P}$, the statement follows.

(b) Choose an arbitrary pair $(b, d) \in \mathcal{P}$ and look for the possible values of $a, c$.

1. Suppose that $b^{5} \equiv d^{5}$, and let $k$ be the number of such pairs $(b, d)$. The value $b$ can be chosen in $p$ different ways. For $b \equiv 0$ only $d=0$ has this property; for the nonzero values of $b$ there are at most 5 possible values for $d$. So we have $k \leq 1+5(p-1)=5 p-4$.

The values $a$ and $c$ must satisfy $a^{2} \equiv c^{2}$, so $a \equiv \pm c$, and there are exactly $2 p-1$ such pairs $(a, c)$.

2. Now suppose $b^{5} \not \equiv d^{5}$. In this case $a$ and $c$ must be distinct. By $(a-c)(a+c)=d^{5}-b^{5}$, the value of $a-c$ uniquely determines $a+c$ and thus $a$ and $c$ as well. Hence, there are $p-1$ suitable pairs $(a, c)$.

Thus, for each of the $k$ pairs $(b, d)$ with $b^{5} \equiv d^{5}$ there are $2 p-1$ pairs $(a, c)$, and for each of the other $p^{2}-k$ pairs $(b, d)$ there are $p-1$ pairs $(a, c)$. Hence,

$$
N=k(2 p-1)+\left(p^{2}-k\right)(p-1)=p^{2}(p-1)+k p \leq p^{2}(p-1)+(5 p-4) p=p\left(p^{2}+4 p-4\right)
$$

To prove the statement of the problem, suppose that $S_{r}=\emptyset$ for some $r \in \mathcal{P}$; obviously $r \not \equiv 0$. Let $T=\left\{x^{10}: x \in \mathcal{P} \backslash\{0\}\right\}$ be the set of nonzero 10th powers modulo $p$. Since each residue class is the 10 th power of at most 10 elements in $\mathcal{P}$, we have $|T| \geq \frac{p-1}{10} \geq 4$ by $p>100$.

For every $t \in T$, we have $S_{t r}=\emptyset$. Indeed, if $(x, y) \in S_{t r}$ and $t \equiv z^{10}$ then

$$
\left(z^{-5} x\right)^{2}+\left(z^{-2} y\right)^{5} \equiv t^{-1}\left(x^{2}+y^{5}\right) \equiv r
$$

so $\left(z^{-5} x, z^{-2} y\right) \in S_{r}$. So, there are at least $\frac{p-1}{10} \geq 4$ empty sets among $S_{1}, \ldots, S_{p-1}$, and there are at most $p-4$ nonzero values among $s_{0}, s_{2}, \ldots, s_{p-1}$. Then by the AM-QM inequality we obtain

$$
N=\sum_{r \in \mathcal{P} \backslash r T} s_{r}^{2} \geq \frac{1}{p-4}\left(\sum_{r \in \mathcal{P} \backslash r T} s_{r}\right)^{2}=\frac{|\mathcal{P} \times \mathcal{P}|^{2}}{p-4}=\frac{p^{4}}{p-4}>p\left(p^{2}+4 p-4\right),
$$

which is impossible by the lemma.
If $5 \nmid p-1$, then all modulo $p$ residue classes are complete fifth powers and the statement is trivial. So assume that $p=10 k+1$ where $k \geq 10$. Let $g$ be a primitive root modulo $p$.

We will use the following facts:

(F1) If some residue class $x$ is not quadratic then $x^{(p-1) / 2} \equiv-1(\bmod p)$.

(F2) For every integer $d$, as a simple corollary of the summation formula for geometric progressions,

$$
\sum_{i=0}^{2 k-1} g^{5 d i} \equiv\left\{\begin{array}{ll}
2 k & \text { if } 2 k \mid d \\
0 & \text { if } 2 k \nmid x d
\end{array} \quad(\bmod p)\right.
$$

Suppose that, contrary to the statement, some modulo $p$ residue class $r$ cannot be expressed as $a^{2}+b^{5}$. Of course $r \not \equiv 0(\bmod p)$. By $(\mathrm{F} 1)$ we have $\left(r-b^{5}\right)^{(p-1) / 2}=\left(r-b^{5}\right)^{5 k} \equiv-1(\bmod p)$ for all residue classes $b$.

For $t=1,2 \ldots, k-1$ consider the sums

$$
S(t)=\sum_{i=0}^{2 k-1}\left(r-g^{5 i}\right)^{5 k} g^{5 t i}
$$

By the indirect assumption and (F2),

$$
S(t)=\sum_{i=0}^{2 k-1}\left(r-\left(g^{i}\right)^{5}\right)^{5 k} g^{5 t i} \equiv \sum_{i=0}^{2 k-1}(-1) g^{5 t i} \equiv-\sum_{i=0}^{2 k-1} g^{5 t i} \equiv 0 \quad(\bmod p)
$$

because $2 k$ cannot divide $t$.

On the other hand, by the binomial theorem,

$$
\begin{aligned}
S(t) & =\sum_{i=0}^{2 k-1}\left(\sum_{j=0}^{5 k}\left(\begin{array}{c}
5 k \\
j
\end{array}\right) r^{5 k-j}\left(-g^{5 i}\right)^{j}\right) g^{5 t i}=\sum_{j=0}^{5 k}(-1)^{j}\left(\begin{array}{c}
5 k \\
j
\end{array}\right) r^{5 k-j}\left(\sum_{i=0}^{2 k-1} g^{5(j+t) i}\right) \equiv \\
& \equiv \sum_{j=0}^{5 k}(-1)^{j}\left(\begin{array}{c}
5 k \\
j
\end{array}\right) r^{5 k-j}\left\{\begin{array}{ll}
2 k & \text { if } 2 k \mid j+t \\
0 & \text { if } 2 k \nmid j+t
\end{array} \quad(\bmod p) .\right.
\end{aligned}
$$

Since $1 \leq j+t<6 k$, the number $2 k$ divides $j+t$ only for $j=2 k-t$ and $j=4 k-t$. Hence,

$$
\begin{gathered}
0 \equiv S(t) \equiv(-1)^{t}\left(\left(\begin{array}{c}
5 k \\
2 k-t
\end{array}\right) r^{3 k+t}+\left(\begin{array}{c}
5 k \\
4 k-t
\end{array}\right) r^{k+t}\right) \cdot 2 k \quad(\bmod p) \\
\left(\begin{array}{c}
5 k \\
2 k-t
\end{array}\right) r^{2 k}+\left(\begin{array}{c}
5 k \\
4 k-t
\end{array}\right) \equiv 0 \quad(\bmod p) .
\end{gathered}
$$

Taking this for $t=1,2$ and eliminating $r$, we get

$$
\begin{aligned}
0 & \equiv\left(\begin{array}{c}
5 k \\
2 k-2
\end{array}\right)\left(\left(\begin{array}{c}
5 k \\
2 k-1
\end{array}\right) r^{2 k}+\left(\begin{array}{c}
5 k \\
4 k-1
\end{array}\right)\right)-\left(\begin{array}{c}
5 k \\
2 k-1
\end{array}\right)\left(\left(\begin{array}{c}
5 k \\
2 k-2
\end{array}\right) r^{2 k}+\left(\begin{array}{c}
5 k \\
4 k-2
\end{array}\right)\right) \\
& =\left(\begin{array}{c}
5 k \\
2 k-2
\end{array}\right)\left(\begin{array}{c}
5 k \\
4 k-1
\end{array}\right)-\left(\begin{array}{c}
5 k \\
2 k-1
\end{array}\right)\left(\begin{array}{c}
5 k \\
4 k-2
\end{array}\right) \\
& =\frac{(5 k) !^{2}}{(2 k-1) !(3 k+2) !(4 k-1) !(k+2) !}((2 k-1)(k+2)-(3 k+2)(4 k-1)) \\
& =\frac{-(5 k) !^{2} \cdot 2 k(5 k+1)}{(2 k-1) !(3 k+2) !(4 k-1) !(k+2) !}(\bmod p) .
\end{aligned}
$$

But in the last expression none of the numbers is divisible by $p=10 k+1$, a contradiction.