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There are infinitely many prime numbers expressible in the form $n^3+1$ where $n$ is a positive integer.

I am not sure whether this is true or not. I tried to prove that it is true via contradition:

Assume via contradiction that there are finite number of prime numbers that can be written in the form $n^3+1$ where $n$ is a positive integer. Then there exists an $N$ such that $N^3+1=P_N$ , where $P_N$ is a prime number and for all $n\ge N$, $n^3+1$ is composite. My idea was to create another prime number greater than $N$ that can be written in this form potentially $(2N)^3+1$ to produce a contradiction. But I am not sure how to construct such a prime number.Can anyone provide hints as to how to approach this problem? I would appreciate hints more than answers.

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    $\begingroup$ Hint: $n^3+1 = (n+1)(n^2-n+1)$. $\endgroup$ – Robert Israel Mar 30 '20 at 22:19
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Hint:

$$n^3 + 1 = (n+1)(n^2-n+1)$$

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    $\begingroup$ If I am understanding correctly $n^3+1=(n+1)(n^2-n+1)$ since this is prime then either $n+1=1$ or $(n^2-n+1)=1$. Note $n+1\neq 1$ since this implies $n=0$ not a positive number and $n^3+1=1$ not a prime so $(n^2-n+1)=1$ which implies $(n-1)=0$ resulting $n=1$ so the only prime that can be written in this form is 2. So the conjecture is false. $\endgroup$ – Noe Vidales Mar 30 '20 at 22:26
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    $\begingroup$ Great stuff, indeed! $\endgroup$ – Victor Mar 30 '20 at 22:29

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