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Source: Own , Mock Thailand Mathematic Olympiad P2

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#1 h Wednesday at 9:07 AM

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Find all $f: \mathbb{R} \to \mathbb{Z^+}$ such that $f(x+f(y))=f(x)+f(y)+1\quad\text{ or }\quad f(x)+f(y)-1$ for all real number $x$ and $y$

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Haris1

#2 h Wednesday at 10:10 AM

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$P(-f(0),0)$ gives $f(a)=1$ for some $a$ .
$P(x,a)$ gives $f(x+1)=f(x)$ or $f(x)+2$
If there is some $b$ such that $f(b+1)=f(b)$
Then $f(b+f(y))$ = $f(b+1+f(y))+2$ or $f(b+1+f(y))$
If at some point we get $f(x+f(y))$ = $f(x+1+f(y))+2$ then taking $y=a$ gives contradiction.
So $f$ -constant for big enough values, so $1=0$ which is not possible.
Otherwise when $f(x+1)=f(x)+2$ for all $x$ .
Doing induction gives $f(n)=c+2n$ for natural numbers $n$
Which gives contradiction when trying natural values for $x$ and $y$ gives contradiction.
So there exist no functions.
Edit: After seeing @down solution i saw that the constant solution $f(x)=1$ worked.

This post has been edited 2 times. Last edited by Haris1, Yesterday at 3:09 PM

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cursed_tangent1434

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cursed_tangent1434

#3 h Yesterday at 6:42 AM

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The answer is $f(x) = 1$ for all $x\in \mathbb{R}$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ be the assertion that $f(x+f(y))=f(x)+f(y)+1 \text{ or } f(x)+f(y)-1$ for real numbers $x$ and $y$ .

Since the range of $f$ is the positive integers the Well-Ordering principle states that there exists a real number $\alpha$ such that $f(x) \ge f(\alpha)$ for all $x \in \mathbb{R}$ . But then, $P(\alpha-f(x),x)$ yields,
$\begin{align*} f\left((\alpha - f(x))+f(x)\right) &= f(\alpha - f(x)) + f(x) \pm 1\\ f(\alpha - f(x)) & = f(\alpha) - f(x) \pm 1 \end{align*}$ However,
$\begin{align*} f(\alpha) \le f(\alpha - f(x)) &= f(\alpha) - f(x) \pm 1\\ f(x) & \le \pm 1 \end{align*}$ which since the range of $f$ is the positive integers implies that we must have $+1$ in $P(\alpha-f(x),x)$ and further, that $f(x) \le 1$ for all $x \in \mathbb{R}$ which immediately allows us to conclude that $f(x)=1$ for all real numbers $x$ as desired.

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