Fermat Number is not Square

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Theorem

There exist no Fermat numbers which are square.


Proof 1

Let $n = 0$.

Then $F_0 = 2^{2^0} + 1 = 3$ is not a square.


Let $n \ge 1$.

Then:

\(\ds F_n\) \(=\) \(\ds 2^{\left({2^n}\right)} + 1\) Definition of Fermat Number
\(\ds \) \(=\) \(\ds \left({2^{\left({2^{n-1} }\right)} }\right)^2 + 1\) ... and so $1$ more than a square

Thus by Zero and One are the only Consecutive Perfect Squares, $2^{\left({2^n}\right)} + 1$ is not square.

$\blacksquare$


Proof 2

Recall the definition of Fermat numbers:

$F_n = 2^{(2^n)}+1$, where $n = 0, 1, 2, \ldots$


Marginal Case

$F_0 = 3$ is not a square.


General Case

It will be shown that Fermat numbers lie between $2$ consecutive squares, thus cannot be itself a square:

\(\ds \left({2^{ \left({2^{n-1} }\right)} }\right)^2\) \(=\) \(\ds 2^{\left({2^n}\right)}\) Index Laws/Product of Indices
\(\ds \) \(<\) \(\ds 2^{\left({2^n}\right)} + 1\) adding a positive term
\(\ds \) \(=\) \(\ds F_n\)
\(\ds \) \(<\) \(\ds 2^{\left({2^n}\right)} + 2^{\left({2^{n-1} }\right) + 1} + 1\) adding another positive term
\(\ds \) \(=\) \(\ds \left({2^{\left({2^{n-1} }\right)} + 1}\right)^2\) Completing the Square

$\blacksquare$


Sources

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