Find a prime $p$ satisfying $p \equiv 1338 \mod 1115$. Are there infinitely many such primes. A little confused about this problem, any help or advice?
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Hint: If $p \equiv 1338 \pmod{1115}$ then $p = 1338+1115n$ for some integer $n$.
Now, note that $1338 = 6 \cdot 223$ and $1115 = 5 \cdot 223$. Hence, $p = 223(6+5n)$.
What does this tell you about $n$?
JimmyK4542
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that there is infinitely many primes numbers $p$ for $n \in Z$ ? – Pasie15 Oct 20 '14 at 17:30
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No, that's not quite right. Using only the condition $p \equiv 1338 \pmod{1115}$ gives us that $p$ is in the form $223(6+5n)$. However, we also know that $p$ is prime. So, can $p = 223(6+5n)$ be the product of two factors greater than $1$? – JimmyK4542 Oct 20 '14 at 18:01
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no by the definition of a prime number – Pasie15 Oct 20 '14 at 18:12
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thus 6 + 5n = 1 ? – Pasie15 Oct 20 '14 at 18:13
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@Pasie15: That's correct. From that equality, you can find out what $n$ has to be, and therefore what the prime number $p$ is using the fact that $p = 1338 + 1115n$. – Michael Albanese Mar 18 '15 at 15:34
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Actually this is easy. Notice that $p \equiv 1338 (mod 1115)$ iif $p-0 \equiv 1338-1115 (mod 1115)$ because $1115 \equiv 0 (mod 1115)$. Also $1338-1115 = 223$ is prime.
Therefore 223 is the prime you're looking for.
Rono
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