Show that there are no $a,b,c,d\in \mathbb{Q}(i)$ such that $e^{2\pi i/5}=a+b\sqrt[4]{2}+c\sqrt{2}+d(\sqrt[4]{2})^{-1}$ Previously in this problem, I found all the subfields of $\mathbb{Q}(\sqrt[4]{2},i)$ with the help of the subgroups of $D_4$ and Galois's Theorem as you can see in the followin...

Show that there are no $a,b,c,d\in \mathbb{Q}(i)$ such that $e^{2\pi i/5}=a+b\sqrt[4]{2}+c\sqrt{2}+d(\sqrt[4]{2})^{-1}$ Previously in this problem, I found all the subfields of $\mathbb{Q}(\sqrt[4]{2},i)$ with the help of the subgroups of $D_4$ and Galois's Theorem as you can see in the followin...

Prove that $\sum_{d|n}\dfrac{1}{d}=\dfrac{\sigma(n)}{n}\ \forall\ n\geq 1, n\in \mathbb Z$ My question and my approach is a lot similar to this question but just a bit short, I want to verify my approach. Here it goes: My Approach: Suppose $\sum_{d|n}\dfrac{1}{d}=\dfrac{1}{d_1}+\dfrac{1}{d_2}+...

Prove that $\sum_{d|n}\dfrac{1}{d}=\dfrac{\sigma(n)}{n}\ \forall\ n\geq 1, n\in \mathbb Z$ My question and my approach is a lot similar to this question but just a bit short, I want to verify my approach. Here it goes: My Approach: Suppose $\sum_{d|n}\dfrac{1}{d}=\dfrac{1}{d_1}+\dfrac{1}{d_2}+...