<s>[INST] Fermat numbers of the form of $b^2$ For n > 1 Let $F_n = 2^{2^n} + 1$ be a fermat number and b = $2^{2^{n - 2}}$ * ($2^{2^{n - 1}}$ - 1 ).
Then $b^2$ $\equiv$ 2 (mod $F_n$)
I tried to square the original expression I got something ugly that I couldn't simplify further.
I got $b^2$ = $2^{2^{n - 1}}$ * ($2^{2^n}$ - $2 * 2^{2^{n - 1}}$ + 1) $\equiv$ - ($2 * 2^{2^n} * 2^{2^{n-1}}$) (mod $F_n$). I stopped here I can't simplify it anymore.
 [/INST] You're almost there! Just notice that $2^{2^n}\equiv -1\pmod{2^{2^n}+1}$ :
$$\begin{align}b^2&=2^{2^{n-1}}(2^{2^n}-2*2^{2^{n-1}}+1)\\&=2^{2^{n-1}}(2^{2^n}+1-2*2^{2^{n-1}})\\&\equiv2^{2^{n-1}}(0-2*2^{2^{n-1}})\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^{n-1}}*2^{2^{n-1}}\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^{n-1}+2^{n-1}}\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^n}\pmod{2^{2^n}+1}\\&\equiv -2*(-1)\pmod{2^{2^n}+1}\end{align}$$
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