Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for any real numbers $x$ and $y:$
\[f(f(x)f(y)) + f(x+y) = f(xy).\]
IMO 2017 Problem 2
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: IMO 2017 Problem 2
Let $P(x,y)$ denote the FE. Note that if $f$ is a solution then $-f$ is also a solution.
Now, $P(2,2)$ imply that $f(f(2)^2)=0$. Let $t=f(2)^2$.
Case 1: $t \neq 1$
$P(x,t)$ implies $f(0)+f(x+t)=f(xt)$, and we can find $x$ so that $x+t=xt$. So $f(0)=0$. Now $P(x,0)$ implies $f(x)=0$ for all real $x$.
Case 2: $t = 1$
So $f(2)=1$ or $-1$. It suffices to deal with the $f(2)=1$ case since $f$ being a solution implies $-f$ is also a solution.
$P(x,1)$ implies $f(0)+f(x+1)=f(x)$, plugging in $1$ here we get that $f(0)=-1$ which implies that $f(x+1)=f(x)+1$ and thus $f(x+n)=f(x)+n$ by induction for integer $n$.
$P(x,0)$ implies that $f(-f(x))+f(x)=-1$ and so $f(-f(x))=-1-f(x)$.
$P(-f(x),0)$ implies that $f(-f(-f(x)))=-1-f(-f(x))$ which implies $f(f(x))=f(x-1)$.
Now assume that $f(k)=0$ but $k \neq 1$. So there exists $r$ with $k+r=kr$. $P(k,r)$ implies $f(0)=0$ which is a contradiction. Note that this implies that $f(k)=n$ for integer $n$ is only possible when $k=n+1$.
Now note that if there exists $p$ and $q$ so that $f(f(p)f(q))=-1$ then $f(p)f(q)=0$ then at least one of $p$ and $q$ equals to $1$.
So if there exists $p$ and $q$ so that $f(pq)-f(p+q)=-1$ or $f(pq+1)-f(p+q)=0$ or $f(pq+1)=f(p+q)$ then at least one of $p$ and $q$ equals to $1$. WLOG $q=1$. But then $pq+1=p+q$.
Now, if $f(a)=f(b)$, then $f(a+n)=f(b+n)$. Note the quadratic $x^2-(a+n)x+(b+n-1)$. The discriminant is $(a+n)^2-4(b+n-1)$ which for large enough $n$ would be positive. So its roots would be real. If the roots are $p$ and $q$, then $a+n=p+q$ and $b+n=pq+1$. Since $f(a+n)=f(b+n)$ we get that $a=b$. So $f$ is injective.
Now, from $f(f(x))=f(x-1)$ we get that $f(x)=x-1$. Plugging it into the FE, it works.
So $0, x-1$ and $1-x$ are the only solutions.
Now, $P(2,2)$ imply that $f(f(2)^2)=0$. Let $t=f(2)^2$.
Case 1: $t \neq 1$
$P(x,t)$ implies $f(0)+f(x+t)=f(xt)$, and we can find $x$ so that $x+t=xt$. So $f(0)=0$. Now $P(x,0)$ implies $f(x)=0$ for all real $x$.
Case 2: $t = 1$
So $f(2)=1$ or $-1$. It suffices to deal with the $f(2)=1$ case since $f$ being a solution implies $-f$ is also a solution.
$P(x,1)$ implies $f(0)+f(x+1)=f(x)$, plugging in $1$ here we get that $f(0)=-1$ which implies that $f(x+1)=f(x)+1$ and thus $f(x+n)=f(x)+n$ by induction for integer $n$.
$P(x,0)$ implies that $f(-f(x))+f(x)=-1$ and so $f(-f(x))=-1-f(x)$.
$P(-f(x),0)$ implies that $f(-f(-f(x)))=-1-f(-f(x))$ which implies $f(f(x))=f(x-1)$.
Now assume that $f(k)=0$ but $k \neq 1$. So there exists $r$ with $k+r=kr$. $P(k,r)$ implies $f(0)=0$ which is a contradiction. Note that this implies that $f(k)=n$ for integer $n$ is only possible when $k=n+1$.
Now note that if there exists $p$ and $q$ so that $f(f(p)f(q))=-1$ then $f(p)f(q)=0$ then at least one of $p$ and $q$ equals to $1$.
So if there exists $p$ and $q$ so that $f(pq)-f(p+q)=-1$ or $f(pq+1)-f(p+q)=0$ or $f(pq+1)=f(p+q)$ then at least one of $p$ and $q$ equals to $1$. WLOG $q=1$. But then $pq+1=p+q$.
Now, if $f(a)=f(b)$, then $f(a+n)=f(b+n)$. Note the quadratic $x^2-(a+n)x+(b+n-1)$. The discriminant is $(a+n)^2-4(b+n-1)$ which for large enough $n$ would be positive. So its roots would be real. If the roots are $p$ and $q$, then $a+n=p+q$ and $b+n=pq+1$. Since $f(a+n)=f(b+n)$ we get that $a=b$. So $f$ is injective.
Now, from $f(f(x))=f(x-1)$ we get that $f(x)=x-1$. Plugging it into the FE, it works.
So $0, x-1$ and $1-x$ are the only solutions.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
-
- Posts:3
- Joined:Thu Aug 27, 2015 12:03 pm
Re: IMO 2017 Problem 2
vaiya ami akvabe krsi aki ans ase but akta problem hochvhehochche. will be pleased if u help
f(f(0)f(0))=f(f(2)f(2))=0
now plugging y=o we get for every x
f(x)=f(0)-f(f(0)f(x))
plugging f(x)=y we get
y=f(0)-f(f(0)y)
now for a certain function f(0) is constant
we assume f(0)=c so
y=c-f(cy)
plugging y=x/c we get
f(x)=f(0)-x/f(0)
now using this ans f(f(2)f(2))=0 we get
f(0)=1 or -1
so f(x)=1-x or x-1
now if f(0)=0
f(x)=f(0)-x/f(0) becomes undefined
so plugging f(0)=0 in f(x)=f(0)-f(f(x)f(0)) we get
f(x)=0
so f(x)=0,x-1or 1-x
could u please tell me where I am wrong??
please don't lauge I beg u
f(f(0)f(0))=f(f(2)f(2))=0
now plugging y=o we get for every x
f(x)=f(0)-f(f(0)f(x))
plugging f(x)=y we get
y=f(0)-f(f(0)y)
now for a certain function f(0) is constant
we assume f(0)=c so
y=c-f(cy)
plugging y=x/c we get
f(x)=f(0)-x/f(0)
now using this ans f(f(2)f(2))=0 we get
f(0)=1 or -1
so f(x)=1-x or x-1
now if f(0)=0
f(x)=f(0)-x/f(0) becomes undefined
so plugging f(0)=0 in f(x)=f(0)-f(f(x)f(0)) we get
f(x)=0
so f(x)=0,x-1or 1-x
could u please tell me where I am wrong??
please don't lauge I beg u
- Atonu Roy Chowdhury
- Posts:64
- Joined:Fri Aug 05, 2016 7:57 pm
- Location:Chittagong, Bangladesh
Re: IMO 2017 Problem 2
]Let $P(x,y)$ denote the assertion $f(f(x)f(y))+f(x+y)=f(xy).$
If $x \neq 1$,
$P(x,\frac{x}{x-1}) \implies f(f(x)f(\frac{x}{x-1}))=0 \text{ [eq.1]} \implies f(0)=0$.
Claim 1: $f(x)=0$
Proof.
$\exists r: f(r)f(\frac{r}{r-1})\neq 1 \forall r \in \mathbb{R}-\{1\}$.
Setting $x=f(r)f(\frac{r}{r-1})$ means $f(0)=0$ [from eq.1] implies that $\boxed{f(x)=0}$, which indeed fits. Q.E.D.
Claim 2: $f$ is injective.
Proof.
Now we can check the opposite case,
$f(x)f(\frac{x}{x-1})=1 \text{ [eq.2] }: x \in \mathbb{R}-\{1\}$.
Eq.1 implies that $f(1)=0 \text{ [denote as eq.3] }.$
Setting $x=0$ means $f(0)^2=1$. Then either $f(0)=1$ or $f(0)=-1 \text{ [eq.4] }$. And we can state that if $f$ is a solution $-f$ is also a solution.
Now by the property of eq.3 and eq.4, $P(x,1) \implies f(x+1)=f(x)+1$ And by induction, indeed, $f(x+n)=f(x)+n \forall n \in \mathbb{N} \text{ [eq.5] }.$
Now assume, $\exists r: f(r)=0, r \neq 1.$
Setting $x=r$ means $0=1$ which clearly implies that $r \neq 1$ is false and it implies that $x=1.$
Thus, $f(x)=0$ if and only if $x=1 \text{ [eq.6] }$.
Again assume, $\exists s: f(s)=-1, s \neq 0.$ Now by eq.5, $f(s)+1=0=f(s+1)$ means $s=0$. Thus, $f(x)=-1$ if and only if $x=0.$
For sake of contradiction, assume $\exists a,b: a\neq b, f(a)=f(b) \forall n \in \mathbb{Z}.$ Which, by eq.5, implies that $f(a+n)=f(b+n) \text{ [eq.6] }$. Given such $a$ and $b$ we may want to find $x$ and $y$ such that $a=x+y$ and $b=xy+1$. If we succeed, we will get our contradiction. From $x+y=a$ and $xy=b-1$, the pair should be the roots of the quadratic equation, $g(u)=u^2-au+(b-1)=0$. For the roots to be real we need a non-negative discriminant: $D=a^2-4(b-1) \geq 0$ which may be not true (i.e. might be negative). But by eq.6, we can convert the discriminant to $D(n)=(a+n)^2-4(b+n-1)$, which is assured of being positive for large $n$.
Now we can conclude from contradiction that $f$ is injective. Q.E.D.
Claim 3: $f(x)=x-1$ and $f(x)=1-x,$ both of which fit.
Proof.
$P(x,1-x)$ from eq.5 means $f(f(x)f(1-x))=f(x(1-x)) \implies f(x)f(1-x)=x(1-x) \text{ [eq.q] }.$
Likewise, $P(x,-x)$ from eq.5 means $f(f(x)f(-x))=f(x^2)+1$. Now by eq.5 $f(f(x)f(-x))=f(-x^2+1) \implies f(x)f(1-x)=x(1-x) \text{ [eq.p] }.$ Now dividing eq.p by eq.q means $f(-x)=-x-1.$
Now setting $x=-x$ implies that (from eq.4) $f(x)=x-1$. Now the fact that "if $f$ is a solution $-f$ is also a solution" provides that also $f(x)=1-x.$ Q.E.D.
If $x \neq 1$,
$P(x,\frac{x}{x-1}) \implies f(f(x)f(\frac{x}{x-1}))=0 \text{ [eq.1]} \implies f(0)=0$.
Claim 1: $f(x)=0$
Proof.
$\exists r: f(r)f(\frac{r}{r-1})\neq 1 \forall r \in \mathbb{R}-\{1\}$.
Setting $x=f(r)f(\frac{r}{r-1})$ means $f(0)=0$ [from eq.1] implies that $\boxed{f(x)=0}$, which indeed fits. Q.E.D.
Claim 2: $f$ is injective.
Proof.
Now we can check the opposite case,
$f(x)f(\frac{x}{x-1})=1 \text{ [eq.2] }: x \in \mathbb{R}-\{1\}$.
Eq.1 implies that $f(1)=0 \text{ [denote as eq.3] }.$
Setting $x=0$ means $f(0)^2=1$. Then either $f(0)=1$ or $f(0)=-1 \text{ [eq.4] }$. And we can state that if $f$ is a solution $-f$ is also a solution.
Now by the property of eq.3 and eq.4, $P(x,1) \implies f(x+1)=f(x)+1$ And by induction, indeed, $f(x+n)=f(x)+n \forall n \in \mathbb{N} \text{ [eq.5] }.$
Now assume, $\exists r: f(r)=0, r \neq 1.$
Setting $x=r$ means $0=1$ which clearly implies that $r \neq 1$ is false and it implies that $x=1.$
Thus, $f(x)=0$ if and only if $x=1 \text{ [eq.6] }$.
Again assume, $\exists s: f(s)=-1, s \neq 0.$ Now by eq.5, $f(s)+1=0=f(s+1)$ means $s=0$. Thus, $f(x)=-1$ if and only if $x=0.$
For sake of contradiction, assume $\exists a,b: a\neq b, f(a)=f(b) \forall n \in \mathbb{Z}.$ Which, by eq.5, implies that $f(a+n)=f(b+n) \text{ [eq.6] }$. Given such $a$ and $b$ we may want to find $x$ and $y$ such that $a=x+y$ and $b=xy+1$. If we succeed, we will get our contradiction. From $x+y=a$ and $xy=b-1$, the pair should be the roots of the quadratic equation, $g(u)=u^2-au+(b-1)=0$. For the roots to be real we need a non-negative discriminant: $D=a^2-4(b-1) \geq 0$ which may be not true (i.e. might be negative). But by eq.6, we can convert the discriminant to $D(n)=(a+n)^2-4(b+n-1)$, which is assured of being positive for large $n$.
Now we can conclude from contradiction that $f$ is injective. Q.E.D.
Claim 3: $f(x)=x-1$ and $f(x)=1-x,$ both of which fit.
Proof.
$P(x,1-x)$ from eq.5 means $f(f(x)f(1-x))=f(x(1-x)) \implies f(x)f(1-x)=x(1-x) \text{ [eq.q] }.$
Likewise, $P(x,-x)$ from eq.5 means $f(f(x)f(-x))=f(x^2)+1$. Now by eq.5 $f(f(x)f(-x))=f(-x^2+1) \implies f(x)f(1-x)=x(1-x) \text{ [eq.p] }.$ Now dividing eq.p by eq.q means $f(-x)=-x-1.$
Now setting $x=-x$ implies that (from eq.4) $f(x)=x-1$. Now the fact that "if $f$ is a solution $-f$ is also a solution" provides that also $f(x)=1-x.$ Q.E.D.