## IMO 2017 Problem 2

Discussion on International Mathematical Olympiad (IMO)
Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

### IMO 2017 Problem 2

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 $y:$

$f(f(x)f(y)) + f(x+y) = f(xy).$
Hammer with tact.

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.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Arman Hassan
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

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm
Location: Chittagong, Bangladesh

### Re: IMO 2017 Problem 2

NO INJECTIVITY!!!
This was freedom. Losing all hope was freedom.