find $f(x)$ when $3f(x-6)-2f(x-9)=x^2-54$ I can easily show that with the assumption $f$ is a polynomial $f(x)=x^2$. But without that assumption how can I prove that $f(x)=x^2$???. I have tried many change of variables $x=u+k$ but to no result. I am lost here
 A: Let $g(x)$ be so that $3g(x)-2g(x-3)=0$. For example, take any periodic $h(x)$ on $[0,1]$ and let $g(x)=h(x/3)\left(\frac23\right)^{x/3}$. Then $g(x)+x^2$ also satisfies your equation. So there are many solutions to your equation.
A: Let $f(x)$ be polynomial of $x$.
Let $n\ge 3,$ be the smallest power of $x$ whose coefficient$(a)$ may be $\ne 0$.
The coefficient of $x^n$ in $3f(x−6)−2f(x−9)$ is $3a-2a=a$
But the coefficient of $x^n$ where $n\ge 3$ in $x^2−54$ is $0\implies a=0$
So, $f(x)$ can not contain any higher powers $(\ge 3)$ of $x.$
So, $f(x)=bx^2+cx+d$
$3f(x−6)−2f(x−9)=x^2−54$
$b(x-h)^2+c(x-h)+d=x^2(b)+x(c-2bh)+(d-ch+bh^2)$
The coefficient of $x^2$ in $3f(x−6)−2f(x−9)$ is $3b-2b=b$
But the coefficient of $x^2$ in $x^2−54$ is $1\implies b=1$
The coefficient of $x$ in $f(x−h)$ is $c-2bh=c-2h$
So, the coefficient of $x$ in $3f(x−6)−2f(x−9)$ is $3\{c-2(-6)\}-2\{c-2(-9)\}=c$
But the coefficient of $x$ in $x^2−54$ is $0\implies c=0$
The constant term  in $f(x−h)$ is $d-ch+bh^2=d+h^2$
The constant term  in $3f(x−6)−2f(x−9$ is $3\{d+(-6)^2\}-2\{d+(-9)^2\}=d-54$
The constant term  in $x^2−54$ is $-54\implies d-54=-54\implies d=0$
So, $f(x)=x^2+0x+0=x^2$
A: The following note - for a general functional equation which is related
Assume that $F=\mathbb{Q}$, $\mathbb{R}$ or
$\mathbb{C}$ and $V$ and $B$ are vector spaces over $F$ and $X$ is a
Banach spaces over $F$.
A function $a:V\rightarrow B$ is said to be additive provided
$a(x+y)=a(x)+a(y)$ for all $x, y\in V$; in this case it is easily
seen that $a(rx) =ra(x)$ for all $x\in V$ and all $r\in \mathbb{Q}$
[1].
If $k\in \mathbb{N}$ and $a:V^{k}\rightarrow B$, then we say that a
is $k$-additive provided it is additive in each variable; we say
that a is symmetric provided
$a(x_{1},x_{2},...,x_{k})=a(y_{1},y_{2},...,y_{k})$ whenever
$x_{1},x_{2},...,x_{k}\in V$ and $(y_{1},y_{2},...,y_{k})$ is a
permutation of $(x_{1},x_{2},...,x_{k})$.
If $k\in \mathbb{N}$ and $a:V^{k}\rightarrow B$ is symmetric and
$k$-additive, let $a^{\ast}(x)=a(x,x,...,x)$ for $x\in V$ is and
note that $a^{\ast}(rx)=r^{k}a(x)$ whenever $x\in V$ and $r\in
\mathbb{Q}$. Such a function $a^{\ast}$ will be called a
monomial function of degree $k$ (assuming $a^{\ast}\neq 0$).
A function $p:V\rightarrow B$ is called a generalized
polynomial} (GP) function of degree $m\in \mathbb{N}$ provided
there exist $a_{0}\in B$ and symmetric $k$-additive functions
$a_{k}:V^{k}\rightarrow B$ (for $1\leq k\leq m$) such that
$$p(x)=a_{0}+\sum_{k=1}^{m}a_{k}^{\ast}(x)\ \ \ for\ all\ x\in V,$$
and $a^{\ast}_{m}\neq 0$. In this case
$$p(rx)=a_{0}+\sum_{k=1}^{m}r^{k}a_{k}^{\ast}(x)\ \ \ for\ all\ x\in V\ and\ r\in \mathbb{Q}.$$
Let $B^{V}$ denote the vector space (over $F$) consisting of all
maps from $V$ into $B$. For $h\in V$ define the linear
difference operator $\Delta_{h}$ on $B^{V}$ by
$$\Delta_{h}f(x)=f(x+h)-f(x)\ \ \ for\ all\ f\in B^{V} and\ x\in V.$$
Notice that these difference operators commute
($\Delta_{h_{1}}\Delta_{h_{2}}=\Delta_{h_{2}}\Delta_{h_{2}}$ for all
$h_{1},h_{2}\in V$) and if $h\in V$ and $n\in \mathbb{N}$, then
$\Delta_{h}^{n}$--the $n$-th iterate of $\Delta_{h}$--satisfies
$$\Delta_{h}^{n}f(x)=\sum_{k=0}^{n}(-1)^{n-k}(_{k}^{n})f(x+kh)\ \ \ for\ f\in B^{V}\ and\  x,h\in V.$$
The following theorem were proved by Mazur and Orlicz [2] and
[3], and in greater generality by Djokovi$\acute{c}$
[4].
Theorem:
If $n\in \mathbb{N}$ and $f:V\rightarrow B$, then the following are
equivalent.
$\Delta_{h}^{n}f(x)=0$ for all $x, h\in V$.
$\Delta_{h_{n}}...\Delta_{h_{1}}f(x)=0$ for all $x,h_{1},...,h_{n}\in V.$
$f$ is a GP function of degree at most $n-1$.
for more information see http://arxiv.org/abs/1210.4975v1
