Let $F|K$ be a field extension and $a \in F$ such that $[K(a):K]$ is odd integer,
then prove that $K(a)=K(a^2)$.
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Since $a$ satisfies a quadratic polynomial with coefficients in $K(a)$ we have $[K(a):K(a^2)]=1$ or $2$. In the former case there's nothing to prove.
In the latter case we get $$ [K(a):K]=[K(a):K(a^2)][K(a^2):K]=2[K(a^2):K]. $$ But this is impossible because $[K(a):K]$ is odd.
Andrea Mori
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How $a$ satisfies a quadratic poly? – yio kk Dec 09 '14 at 19:21
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@yiokk : Since $(a)^2=a^2$, the element $a$ always satisfies $X^2-a^2\in K(a^2)[X]$. – Andrea Mori Dec 09 '14 at 19:24