Fermat Number is not Square/Proof 1
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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.
$\blacksquare$