Find all polynomials `P(x)` such that: `P(x-1).P(x+1)=P(x^2-1)`

 Find all polynomials `P(x)` such that: `P(x-1).P(x+1)=P(x^2-1)`

Solution

Suppose that `\alpha` is a root of `P(x)`  then, `P(\alpha)=0`

Therefore, `P((\alpha+1)-1)=0`

Find the last two digits of : `N=(1!+2!+3!+.......+101!)^101`

Find the last two digits of : `N=(1!+2!+3!+.......+101!)^101` 

Note: This is equivalent to finding `N(mod11)`.

ie: The remainder when dividing `N` by `100`.

Observation: `10!\equiv0(mod100)` Because, `10!=10...5...2`

Therefore, `N\equiv(1!+2!+3!+......+9!)^101(mod100)`

                   `N\equiv(1+2+6+24+20+20+40+20+80)^101(mod100)`

                   `N\equiv13^101(mod100)`