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Prove or disprove :

If $p$ is a prime such that $p\equiv 4 \mod 7,$ then $p \equiv 4\mod 14.$

I think it is a false statement because if $p=11$ then $p\equiv 4 \mod 7$ but $p \not \equiv 4\mod 14.$

Is my counterexample correct?

thanks

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    $\begingroup$ Yes, it is correct. $\endgroup$ – Sahiba Arora Dec 10 '17 at 22:47
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    $\begingroup$ In fact, exactly the opposite is true - if $p$ is prime with $p\equiv 4\pmod 7$ then you never have $p\equiv 4\pmod{14}$. Can you see why? (Hint: what characteristic do all numbers that are $\equiv 4\pmod{14}$ share?) $\endgroup$ – Steven Stadnicki Dec 10 '17 at 22:49
  • $\begingroup$ @dr.rise If you are ok, you can set as solved. Thanks! $\endgroup$ – gimusi Dec 13 '17 at 7:32
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Note that $n\equiv 4 \pmod {14}$ is an even number. Thus it can’t be a prime.

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To add on to gimusi's answer, the proposition could still be correct even given that $n\equiv4\text{ (mod }14)$, but only if $p\equiv4\text{ (mod }7)$ was never true for any prime.

Your counterexample is a valid disproof of the original statement.

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