4
$\begingroup$

For a function $f : \mathbb{N} \times \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ defined by

$f(a,b,c)=(a+b+c)^3+(a+b)^2+a$

I want to show that $f$ is injective.

How can I show this?

I started by assuming $(a,b,c) \neq (d,e,f)$ and $f(a,b,c)=f(d,e,f)$ and seeking a contradiction. However, I am uncertain about the subsequent steps.

Thank you.

$\endgroup$
2
  • 5
    $\begingroup$ Obviously $f(a,b,c)>(a+b+c)^3$. Starter hint: prove that $f(a,b,c)<(a+b+c+1)^3$. What does that say about $(a,b,c)$ and $(d,e,z)$ if $f(a,b,c)=f(d,e,z)$? $\endgroup$
    – Greg Martin
    Commented Dec 10, 2023 at 9:57
  • 2
    $\begingroup$ @Greg Martin I think $a+b+c=d+e+z$, so $(a+b)^2+a=(d+e)^2+d$ holds. And in a similar manner, showing $(a+b)^2 \leq (a+b)^2+a \leq (a+b+1)^2$ implies $a+b=d+e$. Additionally $a=d$, then the conclusion follows. Am I right? Thank you very much. $\endgroup$
    – bel0906
    Commented Dec 10, 2023 at 12:32

0

Reset to default

You must log in to answer this question.