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It says:

Prove by contradiction that any prime number is either of the form of $ab+1$ or $ab+5$.

And this was all. But it seems both belong to $\mathbb{N} \cup \{0\}$. Because otherwise it wouldn't hold for $2$ which is prime.

What the problem says is to say:

$$(p\neq ab+1)\wedge (p\neq ab+5)\implies p\text{ is not prime}$$

I tried to show it would then be written as the product of two numbers and thus not prime. But up to what number should I do that?

Stuck.

AHB
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    Take $a=1$ and $b=p-1$. Don't even need to assume that $p$ is prime. – lulu Jan 10 '17 at 16:11
  • I thought you were asking to prove that it was at least one or the other. How would you write $2$ as $ab+5$? – lulu Jan 10 '17 at 16:14
  • If $p$ is at least $6$ (or if you allow $b<0$) you can take $a=1$, $b=p-5$. But, clearly, these expressions are somewhat silly. I expect you are missing some conditions. – lulu Jan 10 '17 at 16:15
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    Not following. The problem seems entirely trivial...I think something else must be intended. – lulu Jan 10 '17 at 16:17
  • Please trust my word - the problem is meaningless as stated. Check the text again. – Andreas Caranti Jan 10 '17 at 16:29
  • @AndreasCaranti The word to word translation would lead to what I have written in the edit. If it is meaningless, the textbook must have been wrong. – AHB Jan 10 '17 at 16:36
  • @AHB, I trust your word. So there were no quantifications, such as "for some $a, b$..."? – Andreas Caranti Jan 10 '17 at 16:38

1 Answers1

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HINT: Consider modulus 6. A number $p $ taken modulus 6 can only be $0, 1, 2, 3, 4$ or $5$. What happens if $p $ is either $0, 2$ or $4$ mod 6? What if it is $3$ mod 6?

RGS
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  • Why did you choose 6? – AHB Jan 10 '17 at 16:12
  • @AHB some guessing, a bit of intuition and a couple of exercises I once did on modular arithmetic. The point is: if I were to take them modulus 5, the $ab + 5$ would not yield anything meaningful. One tries the next number and luckily it works just fine. – RGS Jan 10 '17 at 16:15
  • It seems you read the question in some way different from @lulu. In a way where this investigation makes sense. I don't see which way. Did you assume any constraints on $a$, $b$ and $p$ beyond those explicitely stated, which motivated your line of thought? – MvG Jan 10 '17 at 16:28