Show that the function $f:\mathbb{Z} \rightarrow \mathbb{2Z}$ by $f(a)=a^3+3a+2$ is not onto

Question

If the function $f:\mathbb{Z} \rightarrow \mathbb{2Z}$ by $f(a)=a^3+3a+2$, then show that $f$ is not onto.

Hint: Show that $f(a)\neq 0$. I have a feeling I have to use the root theorem test, but I don't know how to go about this problem.

Answer 1

To show that $f$ is not onto, you have to show that there exists some element $x \in 2\mathbb{Z}$ such that $f(a) \neq x$ for every $a \in \mathbb{Z}$. Therefore, let $x = 0$ and suppose there exists $a \in \mathbb{Z}$ such that $f(a) = 0$. Then $a^3 + 3a + 2 = 0$. It's easy to see that $f$ is a strictly increasing function. Since our domain is the integers, let's just try a few values of a:

$$ f(-1) = -2$$ $$ f(0) = 2$$

Since $f$ is strictly increasing, any zero of $f$ would have to be in the interval $(-1, 0)$. However, there are no integers in this interval. Therefore there does not exist an integer $a$ such that $f(a) = 0$. Hence, $f$ is not onto.

Answer 2

This is not an algebra problem about roots. Calculate the difference in the value of $f$ at consecutive positive integers: $$f(a+1)-f(a) = (a+1)^3-a^3 +3(a+1-a) = 3a^2 +3a+4 \ge10.$$ This shows the function is increasing and jumps by a quantity of 10 or more. Obviously it cannot cover all even integers. So this function not onlly misses 0 in its image, it misses infinitely many even integers too. (It is possible to interpret this answer as a discretised version of Danny's answer, as divided difference is discrete derivative. But we are able to show more missing numbers in the image).