Let
$2^{q/4}=2^{(4w+1)/4}=2^{1/4} 2^w=X+f$ where q is a prime $X,w\in{N}$ and $0<f<1$.
Since $2^{1/4}$ is irrational $f$ is irrational.
Is there any way to prove that $f^4+4 f^3 X+6 f^2 X^2+4 f X^3$ can be an integer?
Thank you...
Let
$2^{q/4}=2^{(4w+1)/4}=2^{1/4} 2^w=X+f$ where q is a prime $X,w\in{N}$ and $0<f<1$.
Since $2^{1/4}$ is irrational $f$ is irrational.
Is there any way to prove that $f^4+4 f^3 X+6 f^2 X^2+4 f X^3$ can be an integer?
Thank you...
$f^4+4f^3X+6f^2X^2+4fX^3 = (2^{q/4}-X)^4+4(2^{q/4}-X)^3X+6(2^{q/4}-X)^2X^2+4(2^{q/4}-X)X^3=2^q-X^4$