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The question is

It is known that $p$ and $q$ are primes and $n$ is an integer. Denote $(p_1, q_1, n_1), (p_2, q_2, n_2), ..., (p_k, q_k, n_k)$ be the solutions to the equation $pq=n^3-3n+2$. The value of $$\sum_{i=1}^k(p_i+q_i+n_i)$$ is...

This is what I've done so far:

I noticed $n^3-3n+2$ looks pretty inconvenient and decided to factor it out: \begin{equation*} \begin{split} n^3-3n+2 & = n^3-n^2+n^2-3n+2 \\ & = n^2(n-1)+(n-2)(n-1) \\ & = (n-1)(n^2+n-2) \\ & = (n-1)^2(n+2) \end{split} \end{equation*}

This gives me $pq=(n-1)^2(n+2)$ which, in my opinion, looks nice because it gives me some kind of intuition that $pq$ is some number that has the prime factors $(n-1)$ and $(n+2)$. And so, I thought, since $p$ and $q$ must be primes, then $n-1=1$ or $n+2=1$. This gives me $n=2$ and $n=-1$ to try.

Indeed, $n=2$ gives me $pq=4$ and therefore $p=q=2$. Similarly, $n=-1$ gives me $pq=4$ as well and therefore $p=q=2$. So, I have two solutions: $(2, 2, -1)$ and $(2,2,2)$ and therefore $$\sum_{i=1}^2(p_i+q_i+n_i)=(2+2-1)+(2+2+2)=9.$$

Am I missing another solution? If yes, would anyone be willing to point it out to me? Also, any flaw in my thought process that is pointed out would be greatly appreciated. Thanks.

ryan.zcd
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    Not following. In order for $(n-1)^2(n+2)$ to be the product of two primes, you'd need $n-1$ to be $\pm 1$, or $(n+2)=1$. – lulu Nov 16 '22 at 15:44
  • Ah, I forgot about the square root. Thanks. – ryan.zcd Nov 16 '22 at 15:47
  • In any case, when neither term is a unit we get at least two prime factors from the quadratic and one prime factor from the linear term, which is already too much. Thus one must be a unit, which is why you get lulu's comment – Brevan Ellefsen Nov 16 '22 at 15:47
  • More generally see this Key Idea relating the factorizations of polynomials to the factorizations of its values. Here the polynomial is the cubic $(n-1)^2(n+2).\ \ $ – Bill Dubuque Nov 16 '22 at 16:09

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