Prove that there exist infinitely many integers $k$ such that $k$ is not divisible by 5 and $12k+5$ is composite
Prove: there exist infinitely many integers $k$ such that $k$ is not divisible by 5 and $12k+5$ is composite
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elementary-number-theory
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4The question does not make sense. Because the assumption mentions $k$ explicitly, one has to interpret the consequent as being about the same particular $k$ that satisfied the assumption -- however, once $k$ is given there is only _one_ number of the form $12k+5$, whether divisible by 5 or not -- namely $12k+5$ itself. – 2011-09-09
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0@Henning Makholm I might have made a mistake. The question might have been "Prove that if k (integer) is not divisible by 5 then there exists an infinite amount of COMPOSITE numbers of the form 12k + 5. Does that make more sense? – 2011-09-09
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0no. The problem is still that your formulation seems to ask for an infinite number of solutions _for each particular $k$_. But as soon as $k$ is given, $12k+5$ is only one number (which may not even be composite). – 2011-09-09
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0You could validly ask "Prove that there exists infinitely many composite numbers of the form $12k+5$ that are not divisible by $5$", or equivalently "Prove that there exist infinitely many $k$ such that $k$ is not divisible by 5 and $12k+5$ is composite". – 2011-09-09
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0@Henning Makholm What about "For each k which is an integer not divisible by 5 there exists a composite number of the form 12k + 5"? – 2011-09-09
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0That is simply false. Counterexample: 3 is an integer not divisible by 5, but there does not exist any composite number of the form $12\times3+5$. – 2011-09-09
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0@Henning Makholm so I will settle with your version "Prove that there exist infinitely many k such that k is not divisible by 5 and 12k+5 is composite" – 2011-09-09
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0If you are looking for an infinite number of composite numbers of the form $12k+5$ which are not divisible by 5, here are some hints. First note that with $k=1$ the expression gives the answer 17. Can you see a way of adjusting $k$ to get a higher multiple of 17 (try adding the obvious number to $k$) - then an infinite number of multiples of 17. Now some of these may be divisible by 5 - which ones? If you exclude those can you show you still have infinitely many examples left? – 2011-09-09