by DArk0n3 » Sat Apr 20, 2013 6:02 pm
We first start by doing some rearrangements in order to make the polynom easier to work with.
n^8 - n^6 - n^4 + n^2 = n^2(n^6 - n^4 - n^2 +1) = n^2(n^4(n^2 - 1) - (n^2 - 1)) = n^2(n^2 - 1)(n^4 - 1) =
n^2(n - 1)(n +1)(n - 1)(n+1)(n^2 + 1)
Now if n is an odd number this means that n - 1 is an even number and n + 1 is an even number as well. And since n - 1 and n + 1 are two consecutive even numbers this means that one of them is divisible by 4. Then the polynom (n - 1)(n + 1) is divisible by 8. So the whole polynom n^2(n - 1)(n +1)(n - 1)(n+1)(n^2 + 1) is divisible by 64. Since n is odd this means that n^2 is also odd and n^2 + 1 is even. So n^2 + 1 is divisible by 2. So the whole polynom is n^2(n - 1)(n +1)(n - 1)(n+1)(n^2 + 1) is divisible by 128. The numbers n - 1; n; n + 1 are three consecutive numbers, so one of them is divisible by 3. This means that n(n - 1)(n + 1) is divisible by 3 and n^2(n - 1)(n + 1)(n - 1)(n +1) is divisible by 9. So our polynom is divisible by 128*9 = 1152. Now the last part - we have to prove that our polynom is divisible by 5. If n is divisible by 5 then we have no problem - the polynom is divisible by 5. If n is not divisible by 5 then it will be either of the kind 5m + 2 or of the kind 5m + 4, because it is an odd number. If it is of the kind 5m + 2 then n^2 + 1 = (5m + 2)^2 + 1 = 25m^2 +20m + 4 + 1 = 25m^2 + 20m + 5 = 5(5m^2 + 4m + 1), which is divisible by 5. If n is of the kind 5m + 4 then n + 1 will be equal to 5m + 5, which is divisible by 5. So in all cases our polynom is divisible by 5. This means that n^2(n - 1)(n +1)(n - 1)(n+1)(n^2 + 1) is divisible by 1152*5 = 5760