To prove that for a prime $p \equiv 1 \pmod{7}$, there exists some positive integer $m$ such that $m^3 + m^2 - 2m - 1$ is a multiple of $p$, we can proceed as follows:

1. Understanding the Problem:
We need to show that there exists an integer $m$ such that $p$ divides $m^3 + m^2 - 2m - 1$. This means we need to find $m$ such that:
$$m^3 + m^2 - 2m - 1 \equiv 0 \pmod{p}$$

2. Cyclotomic Polynomial Insight:
Since $p \equiv 1 \pmod{7}$, $p$ can be expressed in the form $p = 7k + 1$ for some integer $k$. This implies that $p$ has a primitive 7th root of unity modulo $p$. The 7th cyclotomic polynomial is given by:
$$\Phi_7(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$$
We need to relate this polynomial to our given polynomial $m^3 + m^2 - 2m - 1$.

3. Substitution:
To make the cyclotomic polynomial appear, we use the substitution $m = x + \frac{1}{x}$. This substitution is motivated by the fact that the roots of the cyclotomic polynomial are of the form $e^{2\pi i k / 7}$ for $k = 0, 1, \ldots, 6$.

4. Polynomial Transformation:
Substitute $m = x + \frac{1}{x}$ into the polynomial $m^3 + m^2 - 2m - 1$:
$$m^3 = \left( x + \frac{1}{x} \right)^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}$$
$$m^2 = \left( x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2}$$
$$-2m = -2 \left( x + \frac{1}{x} \right) = -2x - \frac{2}{x}$$
$$-1 = -1$$
Combining these, we get:
$$m^3 + m^2 - 2m - 1 = \left( x^3 + \frac{1}{x^3} + 3x + \frac{3}{x} \right) + \left( x^2 + \frac{1}{x^2} + 2 \right) - \left( 2x + \frac{2}{x} \right) - 1$$
Simplifying, we get:
$$m^3 + m^2 - 2m - 1 = x^3 + \frac{1}{x^3} + x^2 + \frac{1}{x^2} + x + \frac{1}{x} + 1 - 1$$
$$= x^3 + \frac{1}{x^3} + x^2 + \frac{1}{x^2} + x + \frac{1}{x}$$

5. **Existence of $a$ with Order 7 Modulo $p$:**
Since $p \equiv 1 \pmod{7}$, there exists an integer $a$ such that $a$ has order 7 modulo $p$. This means $a^7 \equiv 1 \pmod{p}$ and $a^k \not\equiv 1 \pmod{p}$ for $1 \leq k < 7$.

6. Conclusion:
By choosing $x = a$, where $a$ is a primitive 7th root of unity modulo $p$, we ensure that $x^7 \equiv 1 \pmod{p}$. Therefore, the polynomial $m^3 + m^2 - 2m - 1$ will be zero modulo $p$ for some $m$.

$\blacksquare$