Let `f(x)` be a polynomial function of degree `n` satisfying the condition
`f(x) + f(1/x) = f(x).f(1/x) forall` `x` `in R - {0}`. Then Prove that `f(x) = 1 ± x^n`.

Explanation

Let `f (x) = a_0 + a_1x + ....... + a_nx^n +(a_0 +a_1/x+...+a_n/x^n)`, so
`(a0+a1x+..... + anxn) + = (a0 + a1x +.... + anxn) = (a_0+a_1x+...+a_nx^n)(a_0+a_1/x+...+a_n/x^n)`
Multiply both sides with `x^n`, we get
`(a_0x^n + a_1x^n + 1 +...+ a_nx^(2n))+(a_0x^n+a_1x^(n-1) +...+ a_n)`
`= (a_0 + a_1x+...+a_nx^n) (a_0x^n+a_1x^(n-1) + .... +a_n)`
Equate the coefficients of `x^(2n) , x^(2n - 1), ...... , x^(n + 1)` on both sides, we get
`a_n = a_0 a_n implies a_0 = 1` [ as `a_n ne 0]`
`a^(n - 1) = a_(n - 1) a_0 + a_n a_1 implies a_n a_1 = 0 implies a_1 = 0`
Similarly we get `a_2 = a_3 = ... = a_(n - 1) = 0`
Now equate the coefficient of `x^n` on both sides, we get
`2a_0 = a_0^2 + a_1^2 + ... + a_(n - 1)^2 + a_n^2 implies a_n^2 = 1 implies a_n = +- 1`
Hence `f(x) = 1 +- x^n`