parmenides51 wrote: Find all polynomials $P$ with real coefficients having no repeated roots, such that for any complex number $z$, the equation $zP(z) = 1$ holds if and only if $P(z-1)P(z + 1) = 0$. No constant $P(x)$ fits. It is very easy to get that the only degree $1$ $P(x)$ is $\boxed{P(x)=x}$ If degree is $n\ge 2$: Let $\{z_i\}_{i=1}^n$ the $n$ distinct roots of $P(x)$ $P(z_i)=0$ $\implies$ $P(z_i)P(z_i+2)=0$ $\implies$ $(z_i+1)P(z_i+1)=1$ And so $zP(z)-1$ has at least roots $\{z_i+1\}_{i=1}^n$ and so $zP(z)-1=(\alpha z+a)P(z-1)$ Looking at coefficient of $z^{n+1}$ both sides, we get $\alpha=1$ $P(z_i)=0$ $\implies$ $P(z_i-2)P(z_i)=0$ $\implies$ $(z_i-1)P(z_i-1)=1$ And so $zP(z)-1$ has at least roots $\{z_i-1\}_{i=1}^n$ and so $zP(z)-1=(\beta z+b)P(z+1)$ Looking at coefficient of $z^{n+1}$ both sides, we get $\beta=1$ So $zP(z)-1=(z+a)P(z-1)=(z+b)P(z+1)$ First equality implies $(z+1)P(z+1)-1=(z+a+1)P(z)$ And we also have $(z+b)P(z+1)=zP(z)-1$ And so $(z+b)((z+a+1)P(z)+1)=(z+1)(zP(z)-1)$ Which is $((a+b)z+b(a+1))P(z)=-2z-(b+1)$ Which is impossible since RHS has degree $1$ while LHS is allzero or has degree $\ge 2$ And so no other solution.