Aira's Problem II

By observation of the given equation,

$P(0)=0$

$P(-1)=0$

Hence,

$P(x) = x(x+1)G(x)........................(i)$

Where $G(x)$ is another polynomial function.

Now substitute from equation (i) into the given equation.

$(x-1)x(x+1)G(x-1) = x(x+1)(x-1)G(x)$

Clearly

$G(x-1) = G(x)$ for all real $x$

Since we established that $G(x)$ was a polynomial function, from the above it is clear that it must also be a constant function. Hence $P(x)$ has a degree of $2$

$P(x)=x(x+1)$ So, $(x+1)P(x-1)=(x-1)P(x)$ $(x-1)(x)(x+1)=(x-1)(x)(x+1)$ I Got this formula from observation. So the answer is 2