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If $$F_{n}=2^{2^{n}}+1$$

then compute $F_{10}$ mod $14$,

I thought since $2^{10}=2$ mod $14$

that it would be equal to $5$ mod $14$

but apparently it is $3$.

Anyone can help explain?

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    Why are you taking the mod of the exponent? – 2017-02-23
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    I didnt know you couldnt do that – 2017-02-23

3 Answers 3

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Look at what happens to powers of $2$ modulo $14$. They are: \begin{align*} 2^1&=2\\ 2^2&=4\\ 2^3&=8\\ 2^4&=16\equiv 2\\ 2^5&=32\equiv 4\\ 2^6&=64\equiv 8 \end{align*} Observe that the pattern repeats every $3$ terms. Therefore, the power should be computed modulo $3$, not $14$. You need to compute $2^{10}\pmod 3$ and use that as the exponent on $2$, add $1$, and take that result modulo $14$.

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    Well displayed. (+1) – 2017-02-23
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The order of $2$ mod $7$ is $3$ (i.e. $2^3 = 8 \equiv 1 \mod 7$), so $2^{2^{10}} \mod 7$ depends on $2^{10} \mod 3$, not mod $14$. And then mod $2$, ...

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Note that since $2^4\equiv2\pmod{14}$, it follows that

$$2^n\equiv2^{n\mod3}\pmod{14}$$

From here it's easy to see that $2^{10}\equiv1\pmod3$, so

$$F_{10}=2^{2^{10}}+1\equiv2^1+1=3\pmod{14}$$