Fermat Number is not Square/Proof 1

Theorem

There exist no Fermat numbers which are square.

Proof

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.

$\blacksquare$