# 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.

$\blacksquare$

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $257$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $257$