n=integer<6 if n−1 and n+1 are both prime, and n^2(n^2 +16) is divisible by 720, would the converse be true?
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Following up on @k83f86 s counterexample, $n^2(n^2 + 16)$ is even if and only if $n$ is even, in which case it is automatically a multiple of $16$. $n^2(n^2 + 16)$ is a multiple of $3$ if and only if $n$ is a multiple of $3$ (proof: check the three residue classes mod 3), in which case it is automatically a multiple of $9$. $n^2(n^2 + 16)$ is a multiple of $5$ if and only if $n$ is congruent to $0$, $2$ or $3$ modulo $5$. Thus, by the Chinese remainder theorem, $n^2(n^2 + 16)$ a multiple of $720$ if and only if $n$ is congruent to $0$, $12$ or $18$ modulo $30$. The prime number theorem, or even the much weaker estimate

$$\pi(n) = o(n)$$

implies that there are infinitely many counterexamples.

E: This is assuming the same interpretation of the question as @k83f86. As presently stated, the question makes no sense.

I think the question is: determine whether "n=integer<6" <- "n−1 and n+1 are both prime, and n^2(n^2 +16) is divisible by 720", although I could be wrong.

Your question is not very clear to me.
Perhaps this is what you wanted to ask?
"If 'n' is an integer bigger than 6 and 'n^2 * (n^2 + 16)' is divisible by 720, can we conclude that 'n-1' and 'n+1' are both prime numbers?"

@k83f86 in which case, isn't 48 a counterexample and this is NO? Seems like n = prime#^2 (for n>6) might all fail, but I'm too dumb to know why.

@AlQuinn Yes, in my case this resolves to NO.