Let $ (x-1)^n\mid P(x)$ Prove that $P(x)$ has $n+1$ coefficients not zero Let $ (x-1)^n\mid  P(x)$  Prove that $P(x)$ has  $n+1$ coefficients not zero
It's is 1977 Bulgaria contest, i tried but not succeed
 A: 
I guess the problem must assume $p(x)$ is not the zero polynomial. 

We can assume that $P(0)\neq0$. If it were, we can just divide by a power of $x$. This only shifts the coefficients.
Assume that $p(x)=(x-1)^nq(x)$
Now we take derivative $$p'(x)=n(x-1)^{n-1}q(x)+(x-1)^nq'(x)=(x-1)^{n-1}\left[nq(x)+(x-1)q'(x)\right]$$ 
Observe that $nq(x)+(x-1)q'(x)$ is not the zero polynomial as long as $q(x)$ was not the zero polynomial (look at the leading term). By induction on $n$ we can assume that $p'(x)$ has at least $n-1$ non-zero terms, since we see that it is divisible by $(x-1)^{n-1}$. Since $p'(x)$ only shifts the degrees of non-zero terms of $p(x)$ except the constant, this implies that $p(x)$ has $n+1$ non-zero terms.
We only need to check the first case: If $p(x)\not\equiv0$ is a polynomial divisible by $(x-1)^0$ then it has at least $0$ non-zero terms. This is trivially true. 
Or if you like a less trivial second case: If $p(x)\not\equiv0$ such that $(x-1)$ divides $p(x)$, then $p(x)$ has at least one non-zero term. This one is true since $p(x)$ is not the zero polynomial.
A: We have to assume also that $P$ is a nonzero polynomial. The thesis should be that $Q$ has at least $n+1$ non zero coefficient (otherwise $(x-1)(x-2)=x^2-3x+2$ would be a counterexample).
We can also assume that $P(0)\ne0$, because otherwise we can divide $P$ by the maximum possible power of $x$ and the number of nonzero coefficients is unaffected.
If $n=0$, then the condition is true.
Suppose $(x-1)^n$ divides $P$, with $n>0$. Then the derived polynomial $P'(x)$ is divisible by $(x-1)^{n-1}$, so it has at least $n$ non zero coefficients. So assume
$$
P'(x)=m_1a_1x^{m_1-1}+m_2a_2x^{m_2-1}+\dots+m_ka_kx^{m_k-1}
$$
with $1\le m_1\le m_2\le \dots \le m_k$ and $n\le k$
Then
$$
P(x)=P(0)+a_1x^{m_1}+a_2x^{m_2}+\dots+a_kx^{m_k}
$$
and $P(0)\ne0$.
