Problem 1617

Given a prime $p$, we need to show that there exist two integers $a$ and $b$ such that for all integers $m$, the expression $m^3 + 2017am + b$ is not a multiple of $p$.

We will divide the proof into two cases:

1. **Case (i): $p \neq 2017$**

- Choose $a \equiv 2017^{-1}(p-1) \pmod{p}$. This choice ensures that $2017a \equiv p-1 \pmod{p}$.
- We want to choose $b$ such that $m^3 - m \equiv -b \pmod{p}$ for all $m \in \mathbb{N}$.
- Note that $0^3 - 0 = 0$ and $1^3 - 1 = 0$. Therefore, the set $A = \{ m^3 - m \pmod{p} \mid m = 0, 1, 2, \ldots, p-1 \}$ has at least two repeated zeros.
- Since there are $p$ elements in the set $\{0, 1, 2, \ldots, p-1\}$ and $A$ has at least two repeated zeros, there must exist an integer $c$ in $\{0, 1, 2, \ldots, p-1\}$ that is not in $A$.
- Take $b \equiv -c \pmod{p}$. This ensures that $m^3 + 2017am + b$ is not a multiple of $p$ for any integer $m$.

2. **Case (ii): $p = 2017$

- Claim 1:** The congruence $x^2 \equiv -3 \pmod{2017}$ has a solution in $x$.
- Proof of Claim 1:
$$\left( \frac{-3}{2017} \right) = \left( \frac{-1}{2017} \right) \left( \frac{3}{2017} \right)$$
$$= (-1)^{\frac{2017-1}{2}} \left( \frac{3}{2017} \right)$$
$$= \left( \frac{3}{2017} \right)$$
$$= (-1)^{\frac{2017-1}{2} \cdot \frac{3-1}{2}} \left( \frac{2017}{3} \right)^{-1}$$
$$= \left( \frac{1}{3} \right) = 1$$
Hence, the claim is proven. $\square$

- Choose $x \equiv y \cdot 2^{-1}(-1 + \sqrt{-3}) \pmod{2017}$ where $y \in \{1, 2, 3, \ldots, 2016\}$. We know that $\sqrt{-3} \pmod{2017}$ is meaningful using Claim 1.
- Observe that:
$$x^2 + xy + y^2 \equiv \left( y \cdot 2^{-1}(-1 + \sqrt{-3}) \right)^2 + \left( y \cdot 2^{-1}(-1 + \sqrt{-3}) \right)y + y^2 \pmod{2017}$$
$$\equiv y^2(-2)^{-1}(1 + \sqrt{-3}) + y^2 2^{-1}(-1 + \sqrt{-3}) + y^2 \pmod{2017}$$
$$\equiv y^2 \left( 2^{-1}(-1 - \sqrt{-3} - 1 + \sqrt{-3}) + 1 \right) \pmod{2017}$$
$$\equiv 0 \pmod{2017}$$
- Therefore:
$$(x - y)(x^2 + xy + y^2) \equiv 0 \pmod{2017} \implies x^3 \equiv y^3 \pmod{2017}$$
- Since $x \equiv y \cdot 2^{-1}(-1 + \sqrt{-3}) \pmod{2017}$, it is not hard to verify that $x \not\equiv y \pmod{p}$.
- Consider the set $B = \{ m^3 \pmod{2017} \mid m = 1, 2, 3, \ldots, 2016 \}$. Since we know that there exist $x, y \not\equiv 0 \pmod{2017}$ with $x^3 \equiv y^3 \pmod{2017}$, there must exist an integer $k$ such that $k = 1, 2, 3, \ldots, 2016$ and $k \not\in B$.
- Now, choosing $b = 2017 - k$, we are done. $\blacksquare$