That number is a root of $ t^3-3p^2t=p+p^5 $ This equation has no rational root, since otherwise $ t^3=p Mod p^2 $ which is a contradiction.

How did you get that equation?

A-Thought-Of-God wrote: How did you get that equation? $\sqrt[3]{p} +\sqrt[3]{p^5}=t $ $(\sqrt[3]{p} +\sqrt[3]{p^5})^3=t^3 $ $ p+p^5+3* \sqrt[3]{p} \sqrt{p^5}(\sqrt[3]{p} +\sqrt[3]{p^5})=p+p^5+3p^2t=t^3 $ $ t^3-3p^2t=p+p^5 $

Apart from the obvious approaches used above we can also use some field theory. Let's define $z_1 =\sqrt[3]{p} , \, z_2=\sqrt[3]{p^5}$ , which clearly are algebraic numbers Assuming $z_1 +z_2 \in \mathbb{Q}$ and noting that $z_1 z_2 =p^2 \in \mathbb{Z}$ it follows that a quadratic in $\mathbb{Q}[t]$ can be associated to $z_1 ,z_2$ , which is a contradiction as the degree of the minimal polynomials of $z_1$ and $z_2$ individually are $3$ which clearly $>2$ which is a contradiction $\blacksquare$

Krm wrote: That number is a root of $ t^3-3p^2t=p+p^5 $ This equation has no rational root, since otherwise $ t^3=p Mod p^2 $ which is a contradiction. Why this equation hasn't rational root?

Isn't this trivial because by definition any prime cannot be a cube of another integer?

S.Das93 wrote: Isn't this trivial because by definition any prime cannot be a cube of another integer? Why do we look $(mod p^2)$,l think we can't look because t need not be integer.

The $p^{1/3}$ and $p^{5/3}$ are algebraic integers, being roots of $x^3-p=0$ and $x^3-p^5=0$, so their sum is an algebraic integer. Thus if it is rational, it is an integer. So say $p^{1/3}+p^{5/3}=k$ is an integer. Clearly $k>p$. Let $x=p^{1/3}$. Then $x^5+x-k=0$ and $x^3-p=0$. Since $x^3=p$, $px^2+x-k=0$. Now, $px^3+x^2-kx=0$, and $px^3-p^2=0$. This means that $x^2-kx+p^2=0$, and so $x =\frac{ k \pm \sqrt{k^2-4p^2}}{2}$. Then, $8p=(2x)^3=(k \pm \sqrt{k^2-4p^2})^3$. But then $(4k^2-4p^2) \sqrt{k^2-4p^2}$ is an integer. This means $\sqrt{k^2-4p^2}$ is an integer, and $8p$ is a perfect cube, a contradiction.

Let $t = \sqrt[3]{p} + p\sqrt[3]{p^2}$. Then $t^3 = p + p^5 + 3p^2(\sqrt[3]{p} + p\sqrt[3]{p^2})$, i.e. $t^3 - 3p^2t - p^5 - p = 0$. Suppose, for the sake of contradiction, that $t$ is rational. The rational root theorem for polynomials then implies that $t$ is an integer. Clearly $p$ divides $t^3$ and so (since $p$ is prime) $p$ divides $t$. With $t=kp$ we get $k^3p^3 - 3kp^3 - p^5 = p$, which is impossible since the left-hand side is divisible by $p^3$, while the right-hand side is not.