Fermat Number is not Cube

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Theorem

There exist no Fermat numbers which are cubes.


Proof 1

Direct consequence of Fermat Number is not Perfect Power.

$\blacksquare$


Proof 2

Let $1 + 2^n$ be a Fermat number.

Suppose $a^3 = 1 + 2^n$.

Since $1 + 2^n$ is odd, $a$ must also be odd.

Write $a = 2 m + 1$, where $m > 0$.

Then:

\(\displaystyle 1 + 2^n\) \(=\) \(\displaystyle \paren {2 m + 1}^3\)
\(\displaystyle \) \(=\) \(\displaystyle 8 m^3 + 12 m^2 + 6 m + 1\) Cube of Sum
\(\displaystyle 2^n\) \(=\) \(\displaystyle 8 m^3 + 12 m^2 + 6 m\)
\(\displaystyle \) \(=\) \(\displaystyle 2 m \paren {4 m^2 + 6 m + 3}\)

Since $4 m^2 + 6 m + 3$ is odd and greater than $1$, it is not a power of $2$.

Hence the equation $a^3 = 1 + 2^n$ has no solution in the integers.

Thus there are no Fermat numbers which are cubes.

$\blacksquare$


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