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- Thread starter Pupil
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matt grime

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2*3*..*p_n+1

By construction it is one more than a multiple of any known prime (i.e. 2,..,p_n).

Or, let me put it this way, I have something that is a multiple of 2, and I add one to it. Is the result a multiple of 2?

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2*3*..*p_n+1

By construction it is one more than a multiple of any known prime (i.e. 2,..,p_n).

Or, let me put it this way, I have something that is a multiple of 2, and I add one to it. Is the result a multiple of 2?

No, as far as I can tell by doing a few examples in my head. I can maybe see how 2n + 1 never yields a multiple of 2 since 2n is always even, and adding 1 makes it odd, thereby making it indivisible by 2. Your multiple of 2 example is really simple, so I can see easily how to prove that result, but I don't see how it is known that multiplying all the primes and adding 1 constructs a number such that it isn't a multiple of any known prime.

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HallsofIvy

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Because if you divide kn+ 1 by k you get a quotient of n with remainder 1.

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CRGreathouse

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Because kn is divisible by k (leaving a remainder of 0), so one more than that would leave a remainder of 1.

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where n is an integer.

If p is a factor of k, then

[tex]\frac{kn+1}{p}=n'+\frac{1}{k},[/tex] where n' is an integer.

- #8

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perhaps it would be easier to see it this way: if you cycle through the integers, a multiple of k will appear ever k integers starting with 0 then k then 2k, 3k, etc. If P is the product of all known primes, then it is divisible by every prime, so in order to get a number that's divisible by any factor of P (ie any prime) you need to add at least 2 (the smallest prime) but 1>2 so [tex]2\not| P[/tex]

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