f(m^2)+f(mf(n))=f(m+n)f(m)

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f(m^2)+f(mf(n))=f(m+n)f(m)

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Source: Mexico National Olympiad Mock Exam 2019 P3

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#1 h Oct 15, 2019, 4:07 PM • 2 Y

Y by dangerousliri, Adventure10

Let $\mathbb{Z}$ be the set of integers. Find all functions $f: \mathbb{Z}\rightarrow \mathbb{Z}$ such that, for any two integers $m, n$ , $f(m^2)+f(mf(n))=f(m+n)f(m).$
Proposed by Victor Domínguez and Pablo Valeriano

This post has been edited 1 time. Last edited by plagueis, May 27, 2020, 7:19 AM

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#2 h Oct 15, 2019, 5:50 PM • 5 Y

Y by BlazingMuddy, dangerousliri, AlastorMoody, MarioLuigi8972, Adventure10

plagueis wrote:

Let $\mathbb{Z}$ be the set of integers. Find all functions $f: \mathbb{Z}\rightarrow \mathbb{Z}$ such that, for any two integers $m, n$ , $f(m^2)+f(mf(n))=f(m+n)f(m).$

The only solutions that work are $\boxed{f\equiv 0},\boxed{f\equiv 2},\boxed{f\equiv \text{id}}$ .These clearly work so now we prove these are the only.
Plugging $(0,n)\implies f(0)(f(n)-2)=0\implies f(0)=0$ or $f(n)=2$ .
f(0)=0
f(0) not equal to 0

PS-I think I have flawed in the case $f(1)=0$

This post has been edited 1 time. Last edited by Pluto1708, Oct 15, 2019, 5:59 PM

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#3 h Oct 15, 2019, 6:36 PM • 3 Y

Y by Pluto1708, MarioLuigi8972, Adventure10

That's correct, the induction can only be done backwards. Anyway, you can cover all the integers by choosing some sufficiently small $n$ depending on $m$ , and plug it in the original equation to get $f(m)^2=f(m^2)=0$ for all integers $m$ .

This post has been edited 1 time. Last edited by plagueis, Oct 15, 2019, 6:37 PM

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#4 h Oct 16, 2019, 2:24 AM • 1 Y

Y by Adventure10

Claim : $f(m) = 0 , f(m) = 2 \ and f(m) = m \forall m \in \mathbb{Z}$ are the only solutions.

Let $P(m,n) : f(m^2) + f(mf(n)) = f(m+n)f(m) \forall m,n \in \mathbb{Z} \dots (1)$
$P(0, m ) : 2f(0) = f(m)f(0) \implies f(m) = 2 \ or \ f(0) = 0$ Now observe $f(m) = 2$ is indeed a solution so now we go to the next case.
Case 1 : $f(0)=0$
$P(m,0) : f(m^2) = f(m)^2 \implies f((-m)^2) = f(m^2) = f(-m)^2$ $\implies f(m)^2 = f(-m)^2 \implies f(m) = \pm f(-m) \implies f \ is \ either \ odd \ or \ even$ $P(1,0) : f(1) = f(1)^2 \implies f(1) = 0 \ or \ f(1)= 1$ Case 1.1 : $f(1) = 0$
$P(m,1) : f(m)^2 = f(m+1)f(m) \implies f(m) = 0 \ or \ f(m) = f(m+1)$ Indeed $f(m) = 0 \forall m \in \mathbb{Z}$ is a solution.
Now when $f(m+1) = f(m) \implies f(m) = 0 \forall m \in \mathbb{Z}^+$ ( by simple induction)
And as $f$ is either odd or even we observe that $f(m) = 0 \forall m \in \mathbb{Z}^+ \cup \mathbb{Z}^-$ and as $f(0)=0$ we get -
$f(m) = 0 \forall m \in \mathbb{Z}$ Case 1.2 : $f(1) = 1$
$P(m,1) : f(m)^2 +f(m) = f(m+1)f(m) \implies f(m)= 0 \ or \ f(m+1) = f(m) + 1$ $f(m+1) = f(m) + 1 \implies f(m) = m \forall m \in \mathbb{Z}^+ ( \ by \ simple \ induction )$ Now when $f$ is even $\implies f(m) = | m | \forall m \in \mathbb{Z}$ plugging it back in $eqn(1)$ we observe it isn't a solution.

When $f$ is odd $\implies f(m) = m \forall m \in \mathbb{Z}$ which indeed is a solution.

All solutions of $eqn(1)$ are
$\boxed{f(m) = 2 \forall m \in \mathbb{Z} }$ $\boxed{ f(m) = 0 \forall m \in \mathbb{Z} }$ $\boxed{ f(m) = m \forall m \in \mathbb{Z} }$

This post has been edited 2 times. Last edited by Omega18, Oct 16, 2019, 2:28 AM
Reason: Typos

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#5 h Mar 28, 2021, 11:55 PM

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The only solutions are $f(x)=0$ , $f(x)=2$ and $f(x)=x$ .

Let $P(m,n)$ be the given assertion.
From $P(0,n)$ we have that $2f(0)=f(0)f(n)$ .

Thus we have two cases.

The first case is when $f(0) \neq 0$ , and we get that $f(n)=2$ ., for any $n$ .

The second case is when $f(0)=0$ .
From $P(m,-m)$ we get that $f(m^2)+f(mf(-m))=0$ .
From $P(m,0)$ we get that $f(m)^2=f(m^2)=f(-m)^2$ .
Thus we have that $f(1)^2=f(1)$

Thus we have two cases.

The first case is when $f(1)=1$ .
From $P(m,1)$ , we have that $f(m)\left( f(m)+1 \right)=f(m)f(m+1)$ , which by induction implies that $f(m)=m$ , for every positive integer $n$ .

Now choose $n \geq 0$ and $m \leq 0$ , but choose in such a fashion that $m+n \geq 0$ .
Thus we have that $m^2+f(mn)=(m+n)f(m)$
But now set $m = -1$ , to get that $1+f(-n)=(n-1)f(-1)$ .

From $P(1,-1)$ we get that $f(1)+f(f(-1))=0 \implies f(f(-1))=-f(1)=-1$ .
But notice that $f(-1)^2=1$ :
Now if $f(-1)=1$ , we would get that $f(1)=-1$ , which is impossible, thus $f(-1)=-1$ .
By induction on the upper relation we get that $f(-n)=-n$ , for any positive $n$ .

Thus $f(x)=x$ , for all integers $x$ .

The second case is when $f(1)=0$ , this easily implies that $f(f(n))=0$ and $f(m)^2=f(m+1)f(m)$ , which easily implies, by induction, that $f(x)=0$ .

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