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avatar+692 

Can anyone give a hint here?

 

Given a positive integer $n,$ prove that there exists a positive integer $m > 1000$ such that $m \equiv 7 \pmod{n}$ and $\gcd(m,n) = 1.$

 Aug 15, 2023
 #1
avatar+121 
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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 \(\gcd(m,n) = 1\).

 Aug 15, 2023
 #2
avatar+183 
0

Clearly this statement is false when \(7\vert n\).

 Aug 15, 2023

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