The number of polynomial functions f of degree ≥ 1 satisfying f(x2) = (f (x)) 2 = f (f(x)) is
1
Let degree of f (x) be n.
deg of f(x2) = 2n
deg of ( f(x2)) = 2n
deg of f( f(x) = n2
n2 = 2n iff n = 0 or 2. But n ≥ 1
Degree of f(x) = 2
Since f(x2) = ( f(x2)), if the leading coefficient is a then the coefficient of x4 in the LHS is a and in the RHS is a2.
a = a2 Þ a = 1 ( a = 0 not admissible )
f (x) = x2 + bx + c
f (x2) = x4 + bx2 + c = (f(x))2 = (x2 + bx + c)2
== x4 + 2bx2 + b2x2 + 2cx2 + c2 + 2bcx
Thus Þ 0 = 2b ; ∴ b = 0
b2 + 2c = b Þ 2c = 0 Þ c = 0
∴ The only function is f (x) = x2