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Let $f(x) = ax^3 + bx^2 + cx + d$ be the polynomial with $a\in \mathbb Z.$

Suppose that $f(5) - f(4) =2019.$

Prove that $[f(7) - f (2)]$ is not a prime number.

Thanks all for help!

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$f(5)-f(4) = 61a+9b+c = 2019$.

It is easy to find that $f(7)-f(2) = 335a+45b+5c = 335a+5(2019-61a) = 5(2019+6a)$. Note that $3\mid 2019$ and $3\mid 6a$, we have $2019+6a\ne \pm 1$.

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    $\begingroup$ Since $a\in\mathbb Z,$ so $(2019 + 6a)\neq \pm 1.$ Indeed, if $2019 + 6a = \pm 1$ then $a\in\mathbb Q.$ $\endgroup$ – Success May 10 at 4:16

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