2011/052) prove that x^2+3x+3 is a factor of (x+1)^(n+1) + (x+2)^(2n-1)

for integer n > = 1


This can be proved by induction

n =1 gives (x+1)^(n+1) + (x+2)^(2n-1) = (x+1)^2 + (x+2) = x^2 + 2x + 1 + x + 2 = x^2 + 3x + 3 divisible

so the 1st step is proved

let it be true for n = k.
we need to prove for n= k+ 1

let f(k) = (x+1)^(k+1) + (x+ 2)^(2k- 1)

so f(k+1) = (x+1)^(k+2) + (x+2)^(2k+ 1)
= (x+1)(x+1)^k + (x+2)^2 *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x^2+4x + 4) *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x + 1 + x^2+3x + 3) *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x + 1)(x+2)^(2k-1) + (x^2+3x + 3) *(x+2)^(2k-1)
= (x+1)[(x+1)^k + (x+2)^(2k-1)] + (x^2+3x + 3) *(x+2)^(2k-1)
= (x+1) f(k) + (x^2+3x + 3) *(x+2)^(2k-1)
as both terms are divisible so f(k+1) is divisible

so step of induction is proved

hence proved