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IMO 2017 Problem 2

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IMO 2017 Problem 2

Post Number:#1  Unread postby Thanic Nur Samin » Wed Jul 19, 2017 12:36 am

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 $y:$

\[f(f(x)f(y)) + f(x+y) = f(xy).\]
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Because destroying everything mindlessly isn't cool enough.
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Re: IMO 2017 Problem 2

Post Number:#2  Unread postby Thanic Nur Samin » Wed Jul 19, 2017 1:17 am

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.
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Re: IMO 2017 Problem 2

Post Number:#3  Unread postby Arman Hassan » Mon Jul 24, 2017 8:49 pm

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
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Re: IMO 2017 Problem 2

Post Number:#4  Unread postby Atonu Roy Chowdhury » Thu Jul 27, 2017 12:07 am

NO INJECTIVITY!!!

Notice that, if $f(x)$ is a solution, then $-f(x)$ is also a solution.
Let $P(x,y)$ denotes the assertion.
$P(0,0)$ implies $f(f(0)^2)=0$ . Now if $f(0)=0$ , $P(x,0)$ implies $f(x)=0$ for all $x \in \mathbb{R} $. So, assume $f(0) \neq 0$.

Claim 1: If $f(x)=0$, then $ x=1$
Proof: Assume the contrary. So, $P(x, \frac{x}{x-1})$ implies $f(0)=0$. Contradiction!

So, $f(0)^2=1 \Rightarrow f(0)=1$ or $-1$. WLOG, $f(0)=-1$.

Claim 2: $f(k)=k-1$ for all $k \in \mathbb{Z}$
Proof: $P(x,1) \Rightarrow f(x+1)=f(x)+1$ . Then Induction!

Claim 3: If $a \in R_f$, then $f(a)=a-1$ where $R_f$ denotes the RANGE of $f$. Also, $f(2x)=2f(x)+1$
Proof: $P(x,0) \Rightarrow f(-f(x))=-f(x)-1 \Rightarrow f(f(x))=f(x)-1$ and the other one follows from $P(x,2)$

Claim 4: $f(x)+f(-x)=-2$ and $f(-x)=-f(x+2)$
Proof: $P(x,-1) \Rightarrow f(-x)=f(x)-1+f(-2f(x)) = f(x)+2f(-f(x)) = f(x)-2f(x)-2 = -f(x)-2 = -f(x+2)$

Claim 5: $f(x+y)=f(x)+f(y)+1$
Proof: $f(x)=-f(2-x)$. Then, $f(f(x)f(y))=f(f(2-x)f(2-y)) \Rightarrow f(xy)-f(x+y)=f(xy-2x-2y+4)-f(4-x-y) \Rightarrow f(p)=f(p-2s+4)-f(4-s)+f(s) \Rightarrow f(p)=f(p-2s)+f(2s)+1$ where $p=xy$ and $s=x+y$ and $s^2 \ge 4p$. So, all $x,y$ satisfying $(\frac{y}{2})^2 \ge 4 (x+y)$ or $x+y \le 0$ satisfies $f(x+y)=f(x)+f(y)+1$. Plugging $-x,-y$ into this equation also satisfies the equation. So we can say that, this equation holds for all real $x,y$.
Claim 6: $f(x)f(y)+x+y-xy=1$
Proof: $0=f(f(x)f(y))+f(x+y)-f(xy) =...=f(f(x)f(y)+x+y-xy)$. The result follows.

Putting $y=0$ on claim 6 gives us the result.

So, all such functions are $f(x)=0 , f(x) = x-1 , f(x)=1-x$
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