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I've to study injectivity and surjectivity of $f:\mathbb{Z}\to\mathbb{Z}$, with $f(n)=an^2+bn+c$, in function of $a,b,c\in\mathbb{Z}$.

How can I start?

Here's what I've tried, with injectivity:

From definition, given $n_1,n_2\in\mathbb{Z}$ I have to prove that $f(n_1)=f(n_2)\Rightarrow n_1=n_2$.

Then, $an_1^2+bn_1+c=an_2^2+bn_2+c$

$\Rightarrow a(n_1^2-n_2^2)+b(n_1-n_2)=0$

$\Rightarrow (n_1-n_2)(a(n_1+n_2)+b)=0$

I don't know how to continue.

user26857
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Allonsy
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1 Answers1

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Hint: Parabolas have vertices. If $a > 0$, then $f$ is bounded below, and if $a < 0$, then $f$ is bounded above.

This implies that $f(n) = bn + c$ is linear. Certainly $b \ne 0$.

Now if $b > 1$, the function isn't surjective; note that $c + 1$ isn't in the image of $f$. A similar argument shows that $b \ge -1$, so $b \in \{-1, 1\}$.

It's quite easy to show that the functions $\pm x + c$ are bijections.

  • Yup, I know, when $a\ne0$ then $f$ is definetely not injective. – Allonsy May 22 '14 at 01:16
  • So that implies that $f$ is a linear function; certainly $b \ne 0$ implies injectivity. Now consider $b > 1$ and find a contradiction (and $b < -1$). –  May 22 '14 at 01:17
  • Oh, got it! In $\mathbb{Z}$ there are no inverses. Well, there are only for 1 and -1. So when $b=1$ or $b=-1$, then $f$ is injective, for all $c$. I'm wrong? – Allonsy May 22 '14 at 01:23
  • Right, that's the key idea (except that it's about a failure of surjectivity). I've updated by answer. –  May 22 '14 at 01:23
  • I see. The failure of surjectivity it's on that the function goes "jumping", leaving empty spaces. Well, thank you very much. Now I can continue for myself. – Allonsy May 22 '14 at 01:26
  • Exactly. If you've heard of derivatives, then you can see that the derivative is unbounded - so $f$ leaves (arbitrarily large) gaps between consecutive integers. –  May 22 '14 at 01:28