The required polynomials are L(x) = c (x-3)(x-9)(x-27) (x-81)(x-243) where c is an arbitrary constant, whether real or imaginary).
EXPLANATION:
This problem can be solved by at least two different methods:
METHOD 1:
# Substituting x=1; -242 *L(3) =0 or, L(3) =0, so that (x-3) is a factor of L(x).
Similarly, x=243 gives 243*242*L(243) = 0 or, L(243)=0 so that (x-243) is a factor of L(x).
x=3 gives -240 L(9) = 486 L(3) ; or L(9) = 0 so that (x-9) is a factor of L(x).
x=9 gives -234 L(27) = 243*8*L(9)=0 or, L(27) = 0 so that (x-27) is a factor of L(x).
x=27 gives -216 L(81) = 243*26* L(27)=0 ; or L(81) = 0 so that (x-81) is a factor of L(x).
Hence, L(x) is divisible by each of (x-3), (x-9),(x-27),(x-81) and (x-243).
Let, L(x) = A(n)*(x^n) + A(n-1)*(x^(n-1)) + ------------------+ A(0)( say);
so that,
L(3x) = A(n)*(3^n)*(x^n) + A(n-1)*(3^(n-1))^(x^(n-1)) + ------------------+ A(0);
It can be easily shown that L(3x)/L(x) tends to 3^n as x tends to infinity……….(i)
But, it is given that (x-243)L(3x)= 243(x-1)L(x);
or, L(3x)/L(x) = ( 243 – 243*(1/x)) / (1 – 243 *( 1/x) );
or, L(3x)/L(x) tends to 243 as x tends to infinity………(ii)
(i) and (ii) gives, 3^n =243 or, n=5 and accordingly, L(x) can contain atmost five distinct linear factors.
Since (x-3), (x-9),(x-27),(x-81) and (x-243) are shown to be five linear factors of L(x), it follows that there cannot be any more distinct linear factors and consequently L(x) = c (x-3)(x-9)(x-27) (x-81)(x-243) where c is an arbitrary constant,whether real or imaginary.
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For an alternate solution to this problem, refer to METHOD 2, which was submitted by owl.
METHOD2 (Submitted By owl):
L(x) is a polynomial with (x-243)L(3x)=243(x-1)L(x) for all x.
Substituting x=243 and x=1 gives L(243)=0 and L(3)=0.
Thus, L(x) can be written as
(x-3)(x-243)P(x) where P(x) is a polynomial.
Substituting this into the original identity gives
(x-81)P(3x)=27(x-3)P(x).
Substituting x=81 and x=3 gives P(81)=0 and P(9)=0.
Thus, L(x) can be written as
(x-3)(x-243)(x-81)(x-9)Q(x) where Q(x) is a polynomial.
Substituting this into the original identity gives
(x-27) Q(3 x)=3(x-9) Q(x).
Substituting x=27 and x=9 gives Q(27)=0 and Q(27)=0.
Thus, L(x) can be written as
(x-3)(x-243)(x-81)(x-9)(x-27)R(x) where R(x) is a polynomial.
Substituting this into the original identity gives
R(3 x)=R(x).
The only polynomials that meets this identity are the constant polynomials.
Thus, L(x)=c(x-3)(x-9)(x-27)(x-81)(x-243), where c is any complex number.
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