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.

Natsuki_kobayashi_486 wrote: obviously You are welcome to show us more details of your solution to this certainly entirely trivial problem.

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 $Q(x,y,z)$.)

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.

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.

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

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.

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.

Always 0 conclude answer for 0