Find all functions $f:\mathbb{Z}\rightarrow \mathbb{Z}$, such that for all $a+b+c=0$ holds:
\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]
IMO 2012: Day 2 Problem 4
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Re: IMO 2012: Day 2 Problem 4
Well,I need time for latexing my solution.So only the solution set I am posting.I hope to post the full solution a little later today.Sorry for the inconvenience.
$f(x)=0$ for all integer $x$
$f(x)=ax^{2}$ for any integer $a$.
$f(x)=0,f(y)=a$ for even $x$ and odd $y$
$f(x)=0,f(y)=a,f(z)=4a$ for $x\equiv 0 (mod 4),y\equiv 1 (mod 2),z\equiv 2 (mod 4)$.
I think I enjoyed it!
$f(x)=0$ for all integer $x$
$f(x)=ax^{2}$ for any integer $a$.
$f(x)=0,f(y)=a$ for even $x$ and odd $y$
$f(x)=0,f(y)=a,f(z)=4a$ for $x\equiv 0 (mod 4),y\equiv 1 (mod 2),z\equiv 2 (mod 4)$.
I think I enjoyed it!
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: IMO 2012: Day 2 Problem 4
Really nice. How did you get it?
One one thing is neutral in the universe, that is $0$.
- Tahmid Hasan
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- Joined:Thu Dec 09, 2010 5:34 pm
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Re: IMO 2012: Day 2 Problem 4
Let $P(a,b,c,)$ be the above assertion.
First we find the constant solutions.
Let $f(x)=c \forall x \in \mathbb {Z}$
then $3c^2=6c^2$ or, $f(x)=c=0 \forall x \in \mathbb {Z}$
Now we finfd the non-constant solutions.
$P(0.0.0) \Rightarrow f(0)=0$
$P(a,-a,0) \Rightarrow f(a)=f(-a)$,so $f$ is even.
$P(a+b,-a,-b) \Rightarrow f(a+b)^2+f(a)^2+f(b)^2=2f(a+b)f(a)+2f(a)f(b)+2f(b)f(a+b) \forall a,b \in \mathbb {Z}$
Let $Q(a,b) \Rightarrow f(a+b)^2+f(a)^2+f(b)^2=2f(a+b)f(a)+2f(a)f(b)+2f(b)f(a+b)$
$Q(a,-b) \Rightarrow f(a-b)^2+f(a)^2+f(b)^2=2f(a-b)f(a)+2f(a)f(b)+2f(b)f(a-b)$
$Q(a,b)-Q(a,-b) \Rightarrow (f(a+b)-f(a-b))(f(a+b)+f(a-b)-2f(a)-2f(b))=0$[Let this be R(a,b,)]
so for any $(a,b) \in \mathbb {Z^2}$
$f(a+b)=f(a-b)$.....(1)
or,$f(a+b)+f(a-b)=2f(a)+2f(b)$.....(2)
plugging $b \rightarrow a$ yields $f(2a)=f(0)=0$ or,$f(2a)=4f(a) \forall a \in \mathbb {Z}$
Particularly plugging $a=1\Rightarrow f(2)=0$ or,$f(2)=4f(1)$
Case1:$f(2)=0$
$Q(a,2) \Rightarrow f(a+2)^2+f(a)^2=2f(a+2)f(a)$ or,$f(a+2)=f(a)$
hence $f(a)=0 \forall a$ which are even.
for all odd $a$,$f(a)=f(1)$
notice that in $P(a,b,c)$ there are possible two cases:
1.1.$a,b,c$ are all even.-plugging the values satisfies $P(a,b,c)$
1.2.two are odd,one is even.
now we concentrate on 1.2:
since $P(a,b,c)$ is symmetric,WLOG we assume $a,b$ are odd ,$c$ is even.
Then plugging the values satisfies $P(a,b,c)$
so a solution $f(a)=0 \forall$ even $a$,$f(a)=f(1) \forall$ odd $a$
Case2.$f(2)=4f(1)$
$R(2,1) \Rightarrow (f(3)-f(1))(f(3)-9f(1))=0$
so we get two sunb-cases:
2.1.$f(3)=9f(1)$
we claim $f(a)=a^2f(1) \forall a \in \mathbb {N}$
we prove it by induction.
it is true for $1,2,3$
we assume it is true for $a-1,a$
Then $R(a,1) \Rightarrow f(a+1)=(a+1)^2f(1)$
so $f(a)=a^2f(1) \forall a \in \mathbb {N}$ since the function is even and $f(0)=0$
we conclude $f(a)=a^2f(1) \forall a \in imathbb {Z}$
2.2.f(3)=f(1)
now we claim $f(a)=f(1) \forall a$ which are odd.
$f(a)=0 \forall a \equiv 0(mod 4),f(a)=4f(1) \forall a \equiv 2(mod 4)$
which can be proved using induction as well.
==========================================================
Synthesis of solutions:
1.$f \equiv 0$
2.$f(a)=0 \forall$ even $a$,$f(a)=f(1) \forall$ odd $a$
3.$f(a)=a^2f(1)$
4.$f(a)=0 \forall a \equiv 0(mod 4),f(a)=4f(1) \forall a \equiv 2(mod 4)$
First we find the constant solutions.
Let $f(x)=c \forall x \in \mathbb {Z}$
then $3c^2=6c^2$ or, $f(x)=c=0 \forall x \in \mathbb {Z}$
Now we finfd the non-constant solutions.
$P(0.0.0) \Rightarrow f(0)=0$
$P(a,-a,0) \Rightarrow f(a)=f(-a)$,so $f$ is even.
$P(a+b,-a,-b) \Rightarrow f(a+b)^2+f(a)^2+f(b)^2=2f(a+b)f(a)+2f(a)f(b)+2f(b)f(a+b) \forall a,b \in \mathbb {Z}$
Let $Q(a,b) \Rightarrow f(a+b)^2+f(a)^2+f(b)^2=2f(a+b)f(a)+2f(a)f(b)+2f(b)f(a+b)$
$Q(a,-b) \Rightarrow f(a-b)^2+f(a)^2+f(b)^2=2f(a-b)f(a)+2f(a)f(b)+2f(b)f(a-b)$
$Q(a,b)-Q(a,-b) \Rightarrow (f(a+b)-f(a-b))(f(a+b)+f(a-b)-2f(a)-2f(b))=0$[Let this be R(a,b,)]
so for any $(a,b) \in \mathbb {Z^2}$
$f(a+b)=f(a-b)$.....(1)
or,$f(a+b)+f(a-b)=2f(a)+2f(b)$.....(2)
plugging $b \rightarrow a$ yields $f(2a)=f(0)=0$ or,$f(2a)=4f(a) \forall a \in \mathbb {Z}$
Particularly plugging $a=1\Rightarrow f(2)=0$ or,$f(2)=4f(1)$
Case1:$f(2)=0$
$Q(a,2) \Rightarrow f(a+2)^2+f(a)^2=2f(a+2)f(a)$ or,$f(a+2)=f(a)$
hence $f(a)=0 \forall a$ which are even.
for all odd $a$,$f(a)=f(1)$
notice that in $P(a,b,c)$ there are possible two cases:
1.1.$a,b,c$ are all even.-plugging the values satisfies $P(a,b,c)$
1.2.two are odd,one is even.
now we concentrate on 1.2:
since $P(a,b,c)$ is symmetric,WLOG we assume $a,b$ are odd ,$c$ is even.
Then plugging the values satisfies $P(a,b,c)$
so a solution $f(a)=0 \forall$ even $a$,$f(a)=f(1) \forall$ odd $a$
Case2.$f(2)=4f(1)$
$R(2,1) \Rightarrow (f(3)-f(1))(f(3)-9f(1))=0$
so we get two sunb-cases:
2.1.$f(3)=9f(1)$
we claim $f(a)=a^2f(1) \forall a \in \mathbb {N}$
we prove it by induction.
it is true for $1,2,3$
we assume it is true for $a-1,a$
Then $R(a,1) \Rightarrow f(a+1)=(a+1)^2f(1)$
so $f(a)=a^2f(1) \forall a \in \mathbb {N}$ since the function is even and $f(0)=0$
we conclude $f(a)=a^2f(1) \forall a \in imathbb {Z}$
2.2.f(3)=f(1)
now we claim $f(a)=f(1) \forall a$ which are odd.
$f(a)=0 \forall a \equiv 0(mod 4),f(a)=4f(1) \forall a \equiv 2(mod 4)$
which can be proved using induction as well.
==========================================================
Synthesis of solutions:
1.$f \equiv 0$
2.$f(a)=0 \forall$ even $a$,$f(a)=f(1) \forall$ odd $a$
3.$f(a)=a^2f(1)$
4.$f(a)=0 \forall a \equiv 0(mod 4),f(a)=4f(1) \forall a \equiv 2(mod 4)$
বড় ভালবাসি তোমায়,মা
Re: IMO 2012: Day 2 Problem 4
Here is my solution:
At first we can see easily that the only constant solution of it is:$f(x)\equiv 0$. Now let us set $c=0$.We can show that $f(0)=0$. Since $a+b+c=0$,we can say that $a=-b$. Now,$f(a)^{2}+f(b)^{2}-2f(a)f(b)=(f(a)-f(a))^{2}=0$. Hence the function is even.
Now we again analyze the main equation. We can write it like this:
$(f(a)-f(b))^{2}=f(c)(2f(a)+2f(a)-f(c))$. Now we substitute $a=b=1,c=-2$ and this implies that,$f(2)=0$ or $f(2)=4f(1)$. We have two cases:
$f(2)=0$. If $f(2k)=0$ for any $k$ then $a=2,b=2k,c=-2k-2$ implies that $f(2k+2)=0$. By induction, $f(2n)=0$ for all $n$ and for all odd $x,y$, $f(x)=f(y)$. Thus the solution in this case is $f(x)=0$ for even $x$ and $f(y)=a$ for odd $y$.
$f(2)=4f(1)$. If $f(x)=x^{2}f(1)$ for all $x\leq k$, then, Easy substitutions give us that there arer two possibilities:
$f(k+1)=(k+1)^{2}f(1)$
$f(k+1)=(k-1)^{2}f(1)$.
Case $2.1$:$f(k+1)=(k-1)^{2}f(1)$. Now $a=k+1,b=-k+1,c=-2$ implies that $f(2)=0$ or $f(2)=4(k-1)^{2}f(1)$.
Case $2.1.1$:$f(2)=0$ then $f(1)=0$. By using induction as the previous case,$f(x)=0$ for all $x$.
Case $2.1.2$:$f(2)=4(k-1)^{2}f(1)$. Then $k=2$ and $f(3)=f(1)$ and $f(2)=4f(1)$. Again let us substitute $a=3,b=1,c=-4$. This implies that $f(4)=0 or 4f(2)$. Solving the last two cases,we get a new solution $f(4n)=0,f(2n+1)=a,f(4n+2)=4a$ for all $n$.
Case $2.2$:By induction again,we get a solution from this case: $f(x)=x^{2}f(1)$.
The solution set is given by me before.
P.S.Finding the substitutions was a bit exhausting,but more problemous than that is my net connection. It is high time I changed it for my own good.
At first we can see easily that the only constant solution of it is:$f(x)\equiv 0$. Now let us set $c=0$.We can show that $f(0)=0$. Since $a+b+c=0$,we can say that $a=-b$. Now,$f(a)^{2}+f(b)^{2}-2f(a)f(b)=(f(a)-f(a))^{2}=0$. Hence the function is even.
Now we again analyze the main equation. We can write it like this:
$(f(a)-f(b))^{2}=f(c)(2f(a)+2f(a)-f(c))$. Now we substitute $a=b=1,c=-2$ and this implies that,$f(2)=0$ or $f(2)=4f(1)$. We have two cases:
$f(2)=0$. If $f(2k)=0$ for any $k$ then $a=2,b=2k,c=-2k-2$ implies that $f(2k+2)=0$. By induction, $f(2n)=0$ for all $n$ and for all odd $x,y$, $f(x)=f(y)$. Thus the solution in this case is $f(x)=0$ for even $x$ and $f(y)=a$ for odd $y$.
$f(2)=4f(1)$. If $f(x)=x^{2}f(1)$ for all $x\leq k$, then, Easy substitutions give us that there arer two possibilities:
$f(k+1)=(k+1)^{2}f(1)$
$f(k+1)=(k-1)^{2}f(1)$.
Case $2.1$:$f(k+1)=(k-1)^{2}f(1)$. Now $a=k+1,b=-k+1,c=-2$ implies that $f(2)=0$ or $f(2)=4(k-1)^{2}f(1)$.
Case $2.1.1$:$f(2)=0$ then $f(1)=0$. By using induction as the previous case,$f(x)=0$ for all $x$.
Case $2.1.2$:$f(2)=4(k-1)^{2}f(1)$. Then $k=2$ and $f(3)=f(1)$ and $f(2)=4f(1)$. Again let us substitute $a=3,b=1,c=-4$. This implies that $f(4)=0 or 4f(2)$. Solving the last two cases,we get a new solution $f(4n)=0,f(2n+1)=a,f(4n+2)=4a$ for all $n$.
Case $2.2$:By induction again,we get a solution from this case: $f(x)=x^{2}f(1)$.
The solution set is given by me before.
P.S.Finding the substitutions was a bit exhausting,but more problemous than that is my net connection. It is high time I changed it for my own good.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$