Surjectivity such that f(x) = 0

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Surjectivity such that f(x) = 0

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#1 h Mar 28, 2025, 1:42 AM

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Determine all integer $k$ with this property: For every non-constant function $f : \mathbb{Z} \rightarrow \mathbb{Z}$ that satisfy
$f(f(x) + f(y)) + f(x+y) = 0$ for every integers $x,y$ , there exists an integer $n$ (that depends on choice of $f$ ) such that $f(n) = k$

I need help/fresh approach on whether the family of the solution function is surjective such that there always exist x so f(x) = 0 or not

This post has been edited 1 time. Last edited by KHOMNYO2, Mar 28, 2025, 1:48 AM
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internationalnick123456

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internationalnick123456

#2 h Mar 28, 2025, 2:52 AM • 1 Y

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By considering the function $\begin{cases} f(3n) = 0 \\ f(3n + 1) = u \\ f(3n + 2) = -u\end{cases}$ for some $u\in\mathbb Z$ such that $u\equiv 2\pmod 3$ , we conclude that every $k\neq 0$ does not satisfy the condition.
Now, we will prove that for every function $f:\mathbb Z\to\mathbb Z$ that satisfies $f(f(x) + f(y)) + f(x+y)=0,\forall x,y\in\mathbb Z$ there exists some $n$ such that $f(n) = 0$ .
Indeed, we assume for the sake of contradiction that $0\not\in f(\mathbb Z)$ .
Case 1. $f$ is injective. Then we have $f(f(f(x)) + f(f(y))) = -f(f(x) + f(y)) = f(x+y),\forall x,y\in\mathbb Z$ By injectivity we have $f(f(x))\equiv x$ , thus, $f$ is surjective. (contradiction!)
Case 2. There exists some $u>v$ with $d=u-v$ is minimal such that $f(u) = f(v)$ .
By replacing $y = u$ and $y=v$ , we derive that $f$ is periodic with period $d$ .
Hence, $f(0),f(1),\ldots,f(d-1)$ are pairwise distinct. Since $f(\mathbb Z)$ is symmetric (i.e, $a\in f(\mathbb Z)\Leftrightarrow -a\in f(\mathbb Z)$ ), we must have $d$ is even. Moreover, there exists some $1\leq t\leq d-1$ such that $f(t) = -f(0)$ .
Now, if some integers $x,y$ satisfy $f(x) + f(y) = 0$ , then $f(x+y) = -f(0) = f(t)\Rightarrow x+y\equiv t\pmod d$ Therefore, $f(t - x) + f(x)=0,\forall x\in\mathbb Z\Rightarrow f(2f(x)) = -f(2x) = f(t - 2x),\forall x\in\mathbb Z$
$\Rightarrow 2f(x)\equiv t-2x\pmod d\Rightarrow t$ is even $\Rightarrow f(t/2) = -f(t/2) =0$ , contradiction.
Thus, $k=0$ .

This post has been edited 2 times. Last edited by internationalnick123456, Mar 28, 2025, 2:54 AM

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