need help

To prove this statement, we'll use the Dirichlet's theorem on arithmetic progressions. Dirichlet's theorem states that for any two coprime positive integers $a$ and $d$, there are infinitely many primes in the arithmetic progression $a, a + d, a + 2d, \ldots$.

In our case, we want to find a positive integer $m$ such that $m \equiv 7 \pmod{n}$ and $\gcd(m,n) = 1$. Let $d = n$ and $a = 7$. Since $\gcd(7, n) = 1$ (as $n$ is a positive integer), Dirichlet's theorem guarantees that there are infinitely many primes in the arithmetic progression $7, 7 + n, 7 + 2n, \ldots$.

Let's consider a prime number in this progression that is greater than 1000. Since $n$ is fixed, as we go further along the progression, the numbers will become larger, and eventually, we will find a prime $m > 1000$ that satisfies $m \equiv 7 \pmod{n}$. Moreover, since the prime is greater than 1000 and $n$ is a positive integer, it's guaranteed that $\gcd(m, n) = 1$ because a prime greater than 1000 cannot have any factors in common with $n$.

Therefore, we have shown that for any positive integer $n$, there exists a positive integer $m > 1000$ such that $m \equiv 7 \pmod{n}$ and $\gcd(m,n) = 1$.

Clearly this statement is false when $7\vert n$.