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This is because at least one of them must be divisible by 3.

The outline proof is as follows:

Consider the consecutive odd numbers, 2n+1, 2n+3, 2n+5 ...

Suppose 2n+1 is divisible by 3.

That is 2n+1 = 3k for some integer k, then 2n+7 = (2n+1)+6 = 3k+6 = 3*(k+2) so that 2n+7 is also divisible by 3. So the run stops with 2n+3, 2n+5.

Suppose, instead, that 2n+1 leaves a remainder of 1 when divided by 3.

Then 2n+3 and 2n+9 etc are divisible by 3. Again leaving a run of only two consecutive odd primes.

Finally, suppose that 2n+1 leaves a remainder of 2 when divided by 3. Then 2n+5, 2n+11 etc are divisible by 3 leaving runs of two consecutive odd primes.

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9y ago
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8y ago

Because multiples of three happen with enough frequency to prevent another run like that.

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11y ago

Because multiples of three happen so frequently they prevent another run like that.

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10y ago

Because in later sequences of three consecutive odd numbers, one of them must be divisible by 3.

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Q: Why are 3 5 and 7 are the only three consecutive odd prime numbers?
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