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Let $n,a,d$ be given integers with $(a,d) = 1$.

Let $m = a + q*d$ where q is the product of all primes which divide n but not a.

Prove that $m \equiv a \; (mod \; d)$ and $(m,n) = 1$.

The first part follows from the rules of congruences.

$(m,n) = (a +q*d,n) = \ldots$

My idea would be to apply the Euclid property $(a,b) = (a-bq,b)$ here. However, I didn't manage to go so far with it.

Do you see alternative strategies?

References

Apostol, Introduction to Analytic number theory, Exercise 4.

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user1868607
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1 Answers1

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Let $p$ be a prime number such that $p\mid a+qd$ and that $p\mid n$. There are two possibilities:

  1. $p\mid a$: Then, since $p\mid a+qd$, $p\mid qd$. So, $p\mid q$ or $p\mid d$. Both are impossible: $p\nmid q$ by the definition of $q$ and $p\nmid d$ because $p\mid a$ and $(a,d)=1$.
  2. $p\nmid a$: Then, by the definition of $q$, $p\mid q$. So, $p\mid qd$. So, since we are assuming that $p\mid a+qd$, $p\mid a$, which we are assuming that it does not occur.
José Carlos Santos
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