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Source: Romania EGMO TST 2017 Day 1 P3

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oVlad

#1 h Apr 21, 2025, 1:36 PM

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Determine all functions $f:\mathbb R\to\mathbb R$ such that $f(xy-1)+f(x)f(y)=2xy-1,$ for any real numbers $x{}$ and $y{}.$

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pco

#2 h Apr 21, 2025, 3:54 PM • 1 Y

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oVlad wrote:

Determine all functions $f:\mathbb R\to\mathbb R$ such that $f(xy-1)+f(x)f(y)=2xy-1,$ for any real numbers $x{}$ and $y{}.$

Let $P(x,y)$ be the assertion $f(xy-1)+f(x)f(y)=2xy-1$
Let $c=f(1)$

If $f(0)\ne 0$ , $P(x,0)$ $\implies$ $f(x)$ constant, which is never a solution. So $f(0)=0$

$P(0,0)$ $\implies$ $f(-1)=-1$
$P(1,1)$ $\implies$ $c=\pm 1$
Subtracting $P(x,1)$ from $P(-x,-1)$ , we get $f(-x)=-cf(x)$

Subtracting $P(x,y)$ from $P(xy,1)$ , we get new assertion $Q(x,y)$ : $f(x)f(y)=cf(xy)$
If $f(u)=0$ for some $u\ne 0$ , $Q(x,u)$ implies $f(ux)=0$ $\forall x$ and so $f\equiv 0$ , which is not a solution.
So $f(x)=0$ $\iff$ $x=0$

$Q(x,x)$ implies $\frac{f(x)}c$ is multiplicative and positive $\forall x>0$ and so $g(x)=\ln \frac{f(e^x)}c$ is additive

If $g(x)$ is not linear, its graph is dense in $\mathbb R^2$ and so graph of $f(x)$ is :
Either dense in $\mathbb R_{>0}\times \mathbb R_{>0}$ if $c=1$
Either dense in $\mathbb R_{>0}\times \mathbb R_{<0}$ if $c=-1$

But $P(x,x)$ $\implies$ $f(x^2-1)\le 2x^2-1$ and so contradiction with both cases
So $g(x)$ is linear and $f(x)=cx^a$ $\forall x>0$ for some real $a$
Then $P(2,1)$ implies $c+2^a=3$ and so :

If $c=1$ : $a=1$ and $f(x)=x$ $\forall x\ge 0$ and $f(-x)=-cf(x)=-f(x)$ imply $\boxed{\text{S1 : }f(x)=x\quad\forall x}$ , which indeed fits

If $c=-1$ : $a=2$ and $f(x)=-x^2$ $\forall x\ge 0$ and $f(-x)=-cf(x)=f(x)$ imply $\boxed{\text{S2 : }f(x)=-x^2\quad\forall x}$ , which indeed fits

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#3 h Apr 21, 2025, 6:20 PM

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It is also 2015 Argentina TST P3

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#4 h Apr 21, 2025, 10:44 PM

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https://artofproblemsolving.com/community/c6h2426810p20007653
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