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I tried to follow this example to resolve this exercise, this example to resolve this exercise, however if:

$1+3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}=x$,then

$3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}=(x-1)$, and

$(3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}})^5=3720+2985\cdot 3^{\frac{1}{5}}+2385\cdot3^{\frac{2}{5}}+1905\cdot3^{\frac{3}{5}}+1545\cdot3^{\frac{4}{5}}$

but i cannot factorize the same way since that would result like $1545x+2175+1350\cdot 3^{\frac{1}{5}}+840\cdot3^{\frac{2}{5}}+360\cdot3^{\frac{3}{5}}=(x-1)^5$

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Observe that $$3^{1/5} x = 3^{1/5} + 3^{2/5} + 3^{3/5} + 3^{4/5} + 3 = x + 2,$$ so by rearranging $x = 2/(3^{1/5} - 1)$.

Now if $x \in \mathbb{Q}$, we would get $3^{1/5} \in \mathbb{Q}$, say $3^{1/5} = a/b$ with $a, b \in \mathbb{Z}^+$ coprime. Then $$3b^5 = a^5,$$ forcing $3 \mid a$, whence $3^5 \mid a^5$, so $3 \mid b$ to get enough factors of $3$ on the LHS. But then $a$ and $b$ have a common factor; contradiction! So $3^{1/5} \notin \mathbb{Q}$ and likewise for $x$.

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  • $\begingroup$ Thank you, I was also thinking about the fundamental theorem of algebra to prove the last thing. $\endgroup$ – Emma Wool Nov 1 '17 at 2:04

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