[Guest]

Show that x^2 + x + 1 is a factor of x^(3u) + x^(3v+1) + x^(3w+2) . u,v,w are natural numbers .

[Guest]

this isn't necessarily true.

x^(3*2) +x^(3*5+1) +x^(3*7+2)

x^6 +x^16 +x^23

x^2 +x +1 defiantly isn't a factor of that.

[curr3nt]

x^(3u) mod x^2+x+1 = 1

x^(3v+1) mod x^2+x+1 = x

x^(3w+2) mod x^2+x+1 = x^2

So x^(3u) + x^(3v+1) + x^(3w+2) mod x^2+x+1 = 0 since the remainder of each part would equal x^2+x+1

So if you can prove the first three statements you can prove the problem statement.

[curr3nt]

u = 1 : x^3 - 1 = (x^2 + x + 1) * (x - 1) * (1)

u = 2 : x^6 - 1 = (x^2 + x + 1) * (x - 1) * (x^3 + 1)

u = 3 : x^9 - 1 = (x^2 + x + 1) * (x - 1) * (x^6 + x^3 + 1)

u = n : x^(3n) - 1 = (x^2 + x + 1) * (x - 1) * (x^3(n - 1) + x^3(n - 2) + ... + x^3(1) + x^3(0))

v = 1 : x^4 - x = (x^2 + x + 1) * (x^2 - x) * (1)

v = 2 : x^7 - x = (x^2 + x + 1) * (x^2 - x) * (x^3 + 1)

v = 3 : x^10 - x = (x^2 + x + 1) * (x^2 - x) * (x^6 + x^3 + 1)

v = n : x^(3n+1) - x = (x^2 + x + 1) * (x^2 - x) * (x^3(n - 1) + x^3(n - 2) + ... + x^3(1) + x^3(0))

w = 1 : x^5 - x^2 = (x^2 + x + 1) * (x^3 - x^2) * (1)

w = 2 : x^8 - x^2 = (x^2 + x + 1) * (x^3 - x^2) * (x^3 + 1)

w = 3 : x^11 - x^2 = (x^2 + x + 1) * (x^3 - x^2) * (x^6 + x^3 + 1)

w = n : x^(3n+2) - x^2 = (x^2 + x + 1) * (x^3 - x^2) * (x^3(n - 1) + x^3(n - 2) + ... + x^3(1) + x^3(0))

So...how does one prove those?

I'm pretty sure I'm missing something that makes this proof much easier...