determine all positive integers n such that the polynomial with n+1 terms f(x)=(xpower 4n)+xpower 4(n-1)+....+xpower8+(xpower4)+1 is divisible by g(x)=(xpower 2n)+xpower 2(n-1)+....+xpower4+xpower2)+1.

both f(x) and g(x) are geometric progressions. for f(x) and g(x),a=1,for f(x), r=x⁴, for g(x), it is x²

f(x)/g(x)=((1-x⁴⁽ⁿ⁺¹⁾)/(1-x⁴))((1-x²⁽ⁿ⁺¹⁾)/(1-x²))=(1+x²⁽ⁿ⁺¹⁾)/(1+x²), it is clear from algebra that n+1 is odd, implying that n is any even natural number

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both the num(numerator) and den(denominator) are simple geometric progressions

num/den= ((1-x⁴⁽ⁿ⁺¹⁾)/(1-x⁴))(1-x²)/(1-x²⁽ⁿ⁺¹⁾) = (1+x²⁽ⁿ⁺¹⁾)/(1+x²)

obviously n+1 is odd which implies that n is any even natural number