Find all functions $f:\mathbb {Z}\to\mathbb Z$, satisfy that for any integer ${a}$, ${b}$, ${c}$, $$2f(a^2+b^2+c^2)-2f(ab+bc+ca)=f(a-b)^2+f(b-c)^2+f(c-a)^2$$
Problem
Source: 2023 China TST Problem 22
Tags: algebra, functional equation, function
02.04.2023 17:23
obviously this problem just need to get f(0),f(1),f(-1),f(2),f(-2).and using Mathematical induction we can get the answer is f=0andf(x)=x.
02.04.2023 17:26
Natsuki_kobayashi_486 wrote: obviously You are welcome to show us more details of your solution to this certainly entirely trivial problem.
02.04.2023 19:38
The solutions are zero and identity functions. We prove this via induction and casework on $f(1)$. Let $P(x,y,z)$ denote the given assertion. Firstly, adding $P(a,b,c)$ and $P(-a,-b,-c)$ gives \[2f(\sum a^2) - 2f(\sum ab) = \sum f((a-b)^2) \]so $\sum f(a-b)^2 = \sum f((a-b)^2)$ or we can write this as, $\sum f(x)^2 = \sum f(x^2)$ for variables $x,y,z$ satisfying $x+y+z=0$, further we can also write $\sum f(x)^2 = \sum f(-x)^2$ for these $x,y,z$. (Call this relation $Q(x,y,z)$.)
03.04.2023 17:29
Tintarn wrote: Natsuki_kobayashi_486 wrote: obviously You are welcome to show us more details of your solution to this certainly entirely trivial problem. sorry.because I don't know how to use LaTeX,so I can't give detail here. (I'd already given in other place)but above is all key points in my solution.
03.04.2023 17:53
Natsuki_kobayashi_486 wrote: Tintarn wrote: Natsuki_kobayashi_486 wrote: obviously You are welcome to show us more details of your solution to this certainly entirely trivial problem. sorry.because I don't know how to use LaTeX,so I can't give detail here. (I'd already given in other place)but above is all key points in my solution. A picture of your hand-written solution is also OK.
08.04.2023 06:41
MrOreoJuice wrote: Firstly, adding $P(a,b,c)$ and $P(-a,-b,-c)$ gives \[2f(\sum a^2) - 2f(\sum ab) = \sum f((a-b)^2) \]so $\sum f(a-b)^2 = \sum f((a-b)^2)$ or we can write this as, $\sum f(x)^2 = \sum f(x^2)$ for variables $x,y,z$ satisfying $x+y+z=0$, further we can also write $\sum f(x)^2 = \sum f(-x)^2$ for these $x,y,z$. (Call this relation $Q(x,y,z)$.) I don't see how to get this
08.04.2023 09:11
uh writeup is too tedious, my bad I didn't include the details Firstly $P(a,0,0)$ gives $2f(a^2)=f(a)^2 + f(-a)^2$ so adding $P(a,b,c)$ and $P(-a,-b,-c)$ gives $\text{LHS}=4f(\sum a^2) - 4f(\sum ab)$ and $\text{RHS} = f(a-b)^2 + f(b-c)^2 + f(c-a)^2 + f(b-a)^2 + f(c-b)^2 + f(a-c)^2 = \sum 2f((a-b)^2)$, also $\text{LHS} = 2\sum f(a-b)^2$ because of the original equation.
11.05.2024 01:42
The answers are $f \equiv 0$ and $f \equiv x$. Swapping $b$ and $c$ yields \[f(a-b)^2+f(b-c)^2+f(c-a)^2 = f(a-c)^2+f(c-b)^2+f(b-a)^2.\]So for any $x, y$, we have \[f(x)^2+f(y)^2 + f(-x-y)^2 = f(-x)^2+f(-y)^2 + f(x+y)^2.\]For $g(x) =f(x)^2 - f(-x)^2$, we have $g(x)+g(y)=-g(-x-y)$. Setting $y=0$, $g(x) = -g(-x)$, so $g(x)+g(y) = g(x+y)$ and $g$ is linear over $\mathbb Z$. Now, setting $b=c=0$, we have $2f(a^2) = f(a)^2 + f(-a)^2$. Also, setting $a=b$ and $c=0$, we have $f(2a^2) = f(a^2)$. For $a=1$, these yield $2f(1) = f(1)^2+f(-1)^2$ and $f(2) = 2f(1)$. Setting $a=1$, $b=-1$, $c=0$ yields $2f(2) - 2f(-1) = f(2)^2 + 2f(-1)^2$. Solving these for $f(1)$ yields $f(1) \in \{0, 1\}$. If $f(1) = 1$, then $f(-1) = -1$. It follows that $g \equiv 0$ and $f(-x) = \pm f(x)$. Combining these with previous equations, we get $f(a^2) = f(a)^2$, and the second equation yields $f(2x) = \pm 2f(x)$. The rest is a matter of computation. Letting $c=0$ with $b=1$ and $b=-1$, respectively, \[2f(a) + f(a-1)^2 = 2f(-a) + f(a+1)^2.\]Letting $c=-1$, $b=2$, and setting $a$ and $-a$ respectively yields \[2f(a-2) + f(a-2)^2 + f(a+1)^2 = 2f(-a-2) + f(a+2)^2 + f(a-1)^2.\]Consider the set $S$ of integers $n$ with $f(n) = n$. Note that $1 \in S$. As $f(2) = 2$ and $f(4) = \pm 4$, it follows that $f(3) - f(-3) = 6$, so $f(3) = 3$. Setting $a=2$, we have $f(-2) = 2$. We induct to show that for each $n$, both $n, -n \in S$. Suppose all $n < m$ are in $S$. If $m$ is odd, setting $a=m$ yields $f(m) - f(-m) = 2m$ as $f(m+1)^2 = (m+1)^2$, so $m \in S$. If $m$ is even, then $f(m) = 2 \pm f(m/2)$ or $f(m) = \pm m$. As $f(m+2)^2 = (m+2)^2$, setting $a = m+1$ in $(1)$ yields that $m+1$ and $-(m+1)$ are both in $S$. Then setting $a = m$ yields that $m$ is also itself in $S$. Hence $f \equiv n$ in this case. If $f(1) = 0$, then $f(-1) = 0$ and again $g \equiv 0$. Same as above, $f(a^2) = f(a)^2$ and $f(2a) = \pm 2 f(a)$. Define $S$ to be the set of all $n$ with $f(n) = 0$. Suppose that all $n < m$ are in $S$. Again, if $m$ is odd, set $a = m-1$ in the first display. If $m$ is even, we have $f(m) = \pm 2f(m/2) = 0$. Thus $f \equiv 0$ on $\mathbb Z$ in this case.
05.10.2024 19:30
Always 0 conclude answer for 0