$f:\mathbb{Z}$ $\rightarrow \mathbb{Z}$ defined by $f(x)=mx^3 - nx$, where $m, n \in \mathbb{Z} $ and $m\not|n$.

I am trying to prove whether or not the function is injective/subjective. How could I prove surjectivity for this function without Calculus?


Suppose $f(a) = f(b)$ where $a,b \in \mathbb{Z}$. $ma^3-na = mb^3-nb \iff ma^3-mb^3 = na-nb \iff m(a^3-b^3) = n(a-b) \iff m(a-b)(a^2+ab+b^2)=n(a-b)$.

Case 1) $a-b=0$. Then $a=b$

Case 2) $a-b \not= 0$. Then $m(a^2+ab+b^2)=n \iff a^2+ab+b^2 = \frac{n}{m}$. Contradiction.

Taking case 1), therefore f is injective.


Suppose $f(a)=b$.

  • 2
    $\begingroup$ Good work on the injectivity! I don't think this will necessarily be surjective. Can you think of an example where the image of $f$ only includes even integers? $\endgroup$ Mar 27 at 15:30
  • $\begingroup$ Clearly, in the case where $\gcd(m, n) \neq 1 $, the function isn't surjective. $\endgroup$ Mar 27 at 15:35
  • $\begingroup$ @blacknapkins7, I'm not sure what you're trying to show there exactly. In general $0$ does actually lie in the image of $f$, since $f(0) = 0$. It's true that for $x \ne 0$, we have $f(x) \ne 0$, but this doesn't show $f$ isn't surjective. $\endgroup$ Mar 27 at 16:16
  • $\begingroup$ @IzaakvanDongen An example would be f(2)=8m−2n⟺2(4m−n). Where would I go from there? $\endgroup$ Mar 28 at 16:30

1 Answer 1


I could not say the exact answer to the question, but I don't have enough reputation to add a comment so I'll post Answer. + I first think maybe there must be an condition that $m$, $n$ not zero at the same time.

To prove the surjectivity,

for $f(x) = mx^3 - nx$, put in $ \ n \ $ and $ \ n+1$.


$f(n) = mn^3 - n^2$

$f(n+1) = mn^3+3n^2m+3nm+m-n^2$

So, $f(n+1) - f(n) = 3n^2m+3nm+m = 3m(n^2+n+1)$

When $n$ is $1$ or $-1$, it has the smallest value $1$

Thus, If $m \neq 0$, it is not surjective.

If $m = 0$, $f(x) = -nx$ do the same step as above (compare n and n+1 values) and you can find it is surjective when $n= \pm 1$

Thus, $\begin{cases} \mbox{surjective} & \mbox{if } \ m = 0 \ \& \ n = \pm 1 \\ \mbox{not surjective} & \mbox{Otherwise} \end{cases}$

  • $\begingroup$ I'm undergraduate so, my answer may be wrong. Please feel free to correct me. $\endgroup$
    – sWoo
    Mar 27 at 15:55
  • 1
    $\begingroup$ I don't understand exactly what you're showing here. $f(n)$ and $f(n + 1)$ are just two values of $f$, so how are you concluding anything about surjectivity? I think that the only case where $f$ ends up being surjective is $m = 0$ (as else eventually $f$ grows too fast) and $n = \pm 1$ (as else $1$ doesn't lie in the image). $\endgroup$ Mar 27 at 16:19
  • $\begingroup$ @Izaak van Dongen Oh I made a mistake. f is only surjective when n = $\pm 1$. $\endgroup$
    – sWoo
    Mar 27 at 18:15
  • $\begingroup$ @Izaak van Dongen What I showed is that the value of $f(n+1) - f(n)$ is bigger than $3$ (if $m \neq 0$). For example, if $m=1$, $f(n+1)-f(n)$ is $3(n^2+n+1)$ which means $f(n+1)-f(n) \geq 3$ thus, $\nexists n_0$ such that $f(n_0)=f(n)+1$. $\endgroup$
    – sWoo
    Mar 27 at 18:21
  • $\begingroup$ I don't think that follows. Remember $n$ is fixed, so you've shown that the two specific output values $f(n)$ and $f(n + 1)$ are far apart - this doesn't stop $f$ from being a surjection necessarily. For example, consider the function $f(x)$ given by $f(1) = 10$, $f(10) = 1$ and $f(x) = x$ otherwise. Then $f(1)$ is much bigger than $f(0)$ but $f$ is nevertheless surjective. I think the clearest reason that $f$ fails to be surjective for $m \ne 0$ in this question is the asymptotic behaviour - for large $x$, $\lvert f(x + 1) - f(x) \rvert$ is large, and $f$ is odd, so $f$ can't be surjective. $\endgroup$ Mar 28 at 7:03

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