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Let $a \in - 2 + 5\mathbb{Z}$. Show that the polynomial $f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right) - a$ doesn't have zeroes in $\mathbb{Z}$.

Hint says to prove directly, or by using projection $\mathbb{Z} \to {\mathbb{Z}_5}$ of the polynomial coefficients.

I must admit that I have no idea how to show that, either directly, or via projection. Resulting polynomial is $g\left( x \right) = {x^5} - x - 2$, and it obviously doesn't have zeroes in $\mathbb{Z}$, but I don't know how to, from that fact, justify that $f$ doesn't have any either.

Can this problem be solved within the framework of the field extension theory? Even if not, I'd appreciate a more elaborate hint.

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up vote 5 down vote accepted

Hint: $f(x)+a=x(x-1)(x-2)(x-3)(x-4)\equiv 0(\mod 5)$ for all $x\in\mathbb{Z}$

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There is no solution of $$x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right) + 2 \equiv 0 \pmod 5$$ because: $x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right) \equiv 0 \pmod 5$ for all $x$,

hence there is no solution in integers either.

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