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Source: gghx, p19004309

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#1 h Apr 20, 2025, 6:27 PM

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Find all functions $f:\mathbb Q\to\mathbb Q$ such that for any $x,y\in\mathbb Q$ :
$f(xf(x)+f(x+2y))=f(x)^2+f(y)+y.$

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#2 h Apr 20, 2025, 9:40 PM • 1 Y

Y by jasperE3

Seems like a nice problem! This is my current progress after 30 minutes of work. To be continued... (hopefully)

Denote the given assertion by $P(x, y)$ .

$P(-1, 0)$ gives $f(-1)=0$ . Now comparing $P(-1, \frac{x}{2})$ and $P(0, \frac{x}{2})$ gives that $f(f(x))=f(f(x-1))+f(0)^2$ for all $x\in \mathbb{Q}$ . From simple induction it follows that $f(f(n))=(n+1)f(0)^2+f(0)$ for all integers $n$ .

$P(0, -1)$ gives $-f(0)^2+f(0)=f(0)^2-1$ and this quadratic has solutions $f(0)=1, f(0)=-\frac{1}{2}$ . I'll now split into 2 cases.

Case 1: $f(0)=1$

From the things found before we know $f(f(n))=n+2$ for all integers $n$ and $P(0, n)$ yields $2n+2=1+f(n)+n$ and therefore $f(n)=n+1$ for all integers $n$ .

Case 2: $f(0)=-\frac{1}{2}$

From the things found before we know $f(f(n))=\frac{n-1}{4}$ and $P(0, n)$ yields $f(n)=-\frac{n+1}{2}$ for all integers $n$ .

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#3 h Apr 22, 2025, 4:18 AM

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bump, above poster has the right idea, one more insight is needed (if you haven't seen this idea before)
ideas from the original thread may be useful

This post has been edited 1 time. Last edited by jasperE3, Apr 22, 2025, 4:31 AM

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#4 h Apr 22, 2025, 8:33 PM

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jasperE3 wrote:

bump, above poster has the right idea, one more insight is needed (if you haven't seen this idea before)
ideas from the original thread may be useful

I tried for some more time but didn't really make progress. Only managed to find for all values of the form $\frac{n}{2^k}$ but no clue how to do it for other values. Can you give a hint?

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#5 h Apr 26, 2025, 12:38 AM

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ok ill post my solution

jasperE3 wrote:

Find all functions $f:\mathbb Q\to\mathbb Q$ such that for any $x,y\in\mathbb Q$ :
$f(xf(x)+f(x+2y))=f(x)^2+f(y)+y.$

As in the $\mathbb R\to\mathbb R$ solution, we get $f\left(x+\frac12\right)=f(x)+f(0)^2-\frac12$ and $f(0)\in\left\{-\frac12,1\right\}$ .

Case 1: $f(0)=1$
We have, for $x\in\mathbb Q$ , $f\left(x+\frac12\right)=f(x)+\frac12$ and so $f(x+n)=f(x)+n$ for $n\in\mathbb N$ by induction.
Fix $x\in\mathbb Q$ , let $n>0$ be the denominator of $x+f(x)$ in lowest terms so that $nx+nf(x)$ is an integer (existence guaranteed because $x+f(x)\in\mathbb Q$ ). Then using $P(x+n,y)$ and $P(x,y)$ we have:
$\begin{align*} f(x)^2+f(y)+y+nx+nf(x)+n^2+n&=f(xf(x)+f(x+2y))+nx+nf(x)+n^2+n\\ &=f(xf(x)+f(x+2y)+nx+nf(x)+n^2+n)\\ &=f((x+n)(f(x)+n)+f(x+2y)+n)\\ &=f((x+n)f(x+n)+f(x+2y+n))\\ &=f(x+n)^2+f(y)+y\\ &=f(x)^2+2nf(x)+n^2+f(y)+y \end{align*}$ so simplifying we get $\boxed{f(x)=x+1}$ which satisfies the equation.

Case 2: $f(0)=-\frac12$
Similarly, we have, for $x\in\mathbb Q$ , $f\left(x+\frac12\right)=f(x)-\frac14$ and so $f(x+n)=f(x)-\frac n2$ for $n\in\mathbb N$ by induction.
Fix $x\in\mathbb Q$ , let $n>0$ be the denominator of $-x+2f(x)$ in lowest terms so that $-nx+2nf(x)$ is an integer (existence guaranteed because $-x+2f(x)\in\mathbb Q$ ). Then using $P(x+2n,y)$ and $P(x,y)$ we have:
$\begin{align*} f(x)^2+f(y)+y+\frac12nx-nf(x)+n^2+\frac12n&=f(xf(x)+f(x+2y))+\frac12nx-nf(x)+n^2+\frac12n\\ &=f(xf(x)+f(x+2y)-nx+2nf(x)-2n^2-n)\\ &=f((x+2n)(f(x)-n)+f(x+2y)-n)\\ &=f((x+2n)f(x+2n)+f(x+2y+2n))\\ &=f(x+2n)^2+f(y)+y\\ &=f(x)^2-2nf(x)+n^2+f(y)+y \end{align*}$ so simplifying we get $\boxed{f(x)=\frac{-x-1}2}$ which satisfies the equation.

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