If $f(1), f(0), f(-1)$ are all different then $f$ is constant which is not true since $a\ne 0$... If $f(1)= f(0)$ then $a+b=0$ so $b=-a$, so $$f(c)=f(2a+c)\implies 2a+2c=1$$so for each $a\ne 0$ is solution $$f(x) =ax^2-ax-a+{1\over 2}$$ If $f(-1)= f(0)$ then $a=b$ ... If $f(1)= f(-1)$ then $b=0$ and $a=-2c$ so for each $c\ne 0$ solution is $$f(x)=-2cx^2+c$$

parmenides51 wrote: Find all functions $f(x) = ax^2 + bx + c$, with $a \ne 0$, such that $f(f(1)) = f(f(0)) = f(f(-1))$ . We just have to solve the system $a(a+b+c)^2+b(a+b+c)+c=ac^2+bc+c$ $a(a-b+c)^2+b(a-b+c)+c=ac^2+bc+c$ Which is : $(a+b)(a^2+ab+2ac+b)=0$ $(a-b)(a^2-ab+2ac+b)=0$ And so : Either $a+b=a-b=0$, which means $a=b=0$; impossible Either $a+b=0$ and $a^2-ab+2ac+b=0$ and so $\boxed{(b,c)=\left(-a,\frac {1-2a}2\right)}$ Either $a-b=0$ and $a^2+ab+2ac+b=0$ and so $\boxed{(b,c)=\left(a,\frac {-1-2a}2\right)}$ Either $a^2+ab+2ac+b=a^2-ab+2ac+b=0$ and so $\boxed{(b,c)=\left(0,-\frac a2\right)}$

We can clearly see that $f(1) = a+b+c,f(-1) = a-b+c,f(0) = c$. Therefore, $$f(f(1))=f(f(-1)) \implies f(a+b+c) = f(a-b+c) \implies a(a+b+c)^2+b(a+b+c)+c = a(a-b+c)^2+b(a-b+c)+c \implies$$$$a(a^2+b^2+c^2+2ab+2bc+2ac)+b(a+b+c)+c=a(a^2+b^2+c^2-2ab-2bc+2ac)+b(a-b+c)+c \implies$$$$a(4ab+4bc)+b(2b)=0$$Therefore, either $b=0$ or $a(4a+4c)+2b=0 \implies a(2a+2c)+b=0$. Case 1: $b=0$ If $b=0$, then $f(f(1))=f(a+c)=a(a+c)^2+c$ and $f(f(0))=ac^2+c$. Therefore, $$ac^2+c=a(a^2+2ac+c^2)+c \implies ac^2=a^3+2a^2c+ac^2 \implies a^3+2a^2c=0 \implies a^2+2ac=0 \implies c=-\frac{a}{2}.$$Therefore, $\boxed{(b,c)=\left( 0, -\frac{a}{2}\right)}$. Case 2: $b \ne 0$ We know that $f(f(0))=ac^2+bc+c$ and $f(f(1))=a(a+b+c)^2+b(a+b+c)+c$. Therefore, $$a(a+b+c)^2+b(a+b+c)+c=ac^2+bc+c \implies a(a^2+b^2+2ab+2bc+2ac)+b(a+b)=0 \implies a((a+b)^2+2c(a+b))+b(a+b)=0.$$Case 2a: $a+b=0$ If $a+b=0$ and $-a(-2a+2c)=b$ then, $-a(2a+2c)=-a \implies 2a+2c=1$. Therefore, $(b,c) = \boxed{(b,c)=\left( -a,\frac{1-2a}{2}\right)}$. Case 2b: $a+b \ne 0$ We can divide $a((a+b)^2+2c(a+b))+b(a+b)=0$ by $a+b$ to get that $a(a+b+2c)+b=0$. Additionally, we know that $a(2a+2c)+b=0$. Subtracting them from one-another, $a(-a+b)=0 \implies a=b$. Since $a=b$, $a(2a+2c)+a=0 \implies 2a+2c+1=0 \implies \boxed{(b,c)=\left(a,\frac{-1-2a}{2}\right)}$.

sidenotes