Problem 6

Let n be an integer at least 2. Prove that if

k^2+k+n

is prime for every integer k with

0\le k\le\sqrt{n/3}

, then it is prime for every integer k with

0\le k\le n-2

Step 2 of 5: Step 2

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M\perp\{b+1,\ldots,2b\},\ (2b+1)^2>M\Longrightarrow M\text{ prime}

Detailed analysis

If $M$ is coprime to every integer $b+1,\ldots,2b$ and $(2b+1)^2>M$ , then $M$ is prime. A composite $M$ has a factor $c\le\sqrt M<2b+1$ ; if $c>b$ , it is itself in the interval, while if $c\le b$ , a suitable positive multiple of $c$ lies in $[b+1,2b]$ , a contradiction.