Is $M=\{p\in \mathcal P^3: p(x)=p(x-1), \forall x \in \Bbb R\}$ subspace of $\mathcal P^3$ and if it is, find it's basis. 
Is $M=\{p\in \mathcal P^3: p(x)=p(x-1), \forall x \in \Bbb R\}$ subspace of $\mathcal P^3$ and if it is, find it's basis.

First, I proved that it is subspace:
$\forall \alpha, \beta \in \Bbb R$ ; $p,q\in M$ it holds that $\alpha p+\beta q \in M$
$(\alpha p+\beta q)(x)=\alpha p(x)+\beta q(x)=\alpha p(x-1)+\beta q(x-1)=(\alpha p+\beta q)(x-1)$
Ok now I want to find a basis.
\begin{align}
ax^3+bx^2+cx+d & = a(x^3-3x^2+3x-1)+b(x^2-2x+1)+c(x-1)+d \\
&= ax^3-3ax^x+3ax-3a+bx^2-2bx+b+cx-c+d\\
\end{align}
What I get is:
$$0=-3ax^2+3ax-3a-2bx+b-c$$
$$3ax^2-x(3a-2b)+3a+b-c=0$$
Now what? I don't know if this is the easiest way of finding the base, but I just want to learn how to solve this type of problem in this, more explicit, way.
 A: You have $M=\mathbb R$, so any number will be a basis. For instance $\{1\}$ is a basis. Or $\{7\sqrt2\}$, or any other number. 
If $p$ is a degree three polynomial with $p(x)=p(x-1)$, we have 
\begin{align}
p(x)=ax^3+bx^2+cx+d &= a(x-1)^3+b(x-1)^2+c(x-1)+d \\ \ \\
&= a(x^3-3x^2+3x-1)+b(x^2-2x+1)+c(x-1)+d\\ \ \\
&=ax^3 +(-3a+b)x^2+(3a-2b+c)x-a+b-c+d.
\end{align}
From $-3a+b=b$, we get $a=0$. From $3a-2b+c=c$, we get $b=0$. And from $-c+d=d$, we get $c=0$. So $p(x)=d$ is constant. 
Here's an even easier way. From $p(x)=p(x-1)$, we get that $p'(x)=p'(x-1)$, $p''(x)=p''(x-1)$, etc. for all derivatives. Comparing the second derivatives, we get $$ 6ax=6a(x-1)$$ for all $x$, which implies $a=0$. Now comparing second derivatives, $2bx=2b(x-1),$ implying that $b=0$. In a similar fashion it follows that $c=0$. 
A: A polynomial is equal to $0$ if and only if all of its coefficients are equal to $0$. From your equation
$$3ax^2-x(3a-2b)+3a+b-c=0$$
 we obtain:
$$3a = 0$$
$$3a - 2b = 0$$
$$3a + b - c = 0$$
Hence, $a = b = c = 0$. We have no condition on $d$, so $d \in \mathbb{R}$.
So, $M$ is the space of constant polynomials. Its basis is for example $\{1\}$.
