Fermat Number is not Square/Proof 2

Theorem

There exist no Fermat numbers which are square.

Recall the definition of Fermat numbers:

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

$F_0 = 3$ is not a square.

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

$\ds \left({2^{ \left({2^{n-1} }\right)} }\right)^2$

$=$

$\ds 2^{\left({2^n}\right)}$

$\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 $

$=$

$\ds \left({2^{\left({2^{n-1} }\right)} + 1}\right)^2$

$\blacksquare$