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First, we need to realize that for any whole number n, one of n+1, n, or n-1 is divisible by three.
Also, if n is prime, then n is not divisible by three (unless n=3), so either n+1 or n-1 is divisible by three. In that case,
(n+1)(n-1)=n²-1
and
n²-1+9=n²+8
will also be divisible by three.
Therefore, the only time that n and n²+8 will both be prime is if n is equal to three. In this situation, x³+16=43 is indeed prime, so we have proven the original supposition. |
