If $y$, $x$,are natural numbers, and $n$ is a prime number, $y = x + n$, $y>x>n$, and $y$ and $x$ are not coprime, is it true that $n$ is a divisor of both $x$ and $y$? If so could you please start off the proof and point me in the general direction of how to prove it, please don't show me the full proof, I want to solve it myself.
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$\begingroup$ Hint: $6=2+4$ . $\endgroup$– UncountableApr 25, 2015 at 19:13
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$\begingroup$ @Uncountable I know that, that's why I asked this question. $\endgroup$– tox123Apr 25, 2015 at 19:14
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$\begingroup$ The reason I said it is because I think this suffices as a counterexample. Correct me if I am wrong. $\endgroup$– UncountableApr 25, 2015 at 19:15
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$\begingroup$ If you know that how can you still ask whether $4\mid 6$ and $4\mid 2$ are true? $\endgroup$– Hagen von EitzenApr 25, 2015 at 19:15
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$\begingroup$ Oh dang, I forgot a specification. $\endgroup$– tox123Apr 25, 2015 at 19:16
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1 Answer
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You can even say that any divisor of both $x$ and $y$ is also a divisor of $n$, hence $\gcd(x,y)=p$.
If $x$ and $y$ are coprime this is not necessarily true. Counter-example: twin primes differ by $2$, a prime number.
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$\begingroup$ God, so many counter-examples, so many specifications needed to avoid them. $\endgroup$– tox123Apr 25, 2015 at 19:23
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$\begingroup$ Oh! The post was modified when I was writing my answer, and I didn't check before posting. See my updated answer. $\endgroup$– BernardApr 25, 2015 at 19:34
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$\begingroup$ I don't know why my answer was downvoted: it corresponded to the first state of the O.P.'s question and was perfectly correct. I have modified it since then. $\endgroup$– BernardApr 25, 2015 at 19:36
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$\begingroup$ I don't feel like make more specifications in my question, so if x, and y are composite, is it correct, if so how would I prove that? $\endgroup$– tox123Apr 25, 2015 at 19:40
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$\begingroup$ I don't quite understand what are the specifications finally: $x$ and $y$ are composite and not coprime? $\endgroup$– BernardApr 25, 2015 at 19:46