compute the remainder when 2^(2^17) + 1 = 19

that + 1 is really throwing me off... I don't know how to deal with it.

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- Feb 28th 2009, 08:58 AMCoda202fermat's little thm
compute the remainder when 2^(2^17) + 1 = 19

that + 1 is really throwing me off... I don't know how to deal with it. - Feb 28th 2009, 10:37 AMThePerfectHacker
I assume you want to find the remainder of $\displaystyle 2^{2^{17}}+1$ modulo $\displaystyle 19$.

By the division algorithm we can write $\displaystyle 2^{17} = 18q + r$ where $\displaystyle 0\leq r < 18$.

But then, $\displaystyle 2^{2^{17}} = 2^{18q + r} = \left( 2^{18} \right)^q \cdot 2^r \equiv 2^r (\bmod 19)$.

To find the remainder $\displaystyle r$ we need to simplify $\displaystyle 2^{17}$ modulo $\displaystyle 18$.

First, $\displaystyle 2^{\phi(9)} \equiv 1(\bmod 9) \implies 2^6\equiv 1(\bmod 9) \implies 4\cdot 2^{16} \equiv 1(\bmod 9)$.

Multiply both sides by seven to get, $\displaystyle 2^{16}\equiv 7(\bmod 9) \implies 2^{17}\equiv 14(\bmod 18)$.

Therefore, $\displaystyle 2^{2^{17}}\equiv 2^{14} (\bmod 19)$.

Now, $\displaystyle 16\cdot 2^{14}\equiv 1(\bmod 19)$ by Fermat's little theorem.

Since $\displaystyle 16\cdot 6\equiv 1(\bmod 19)$ we see that $\displaystyle 2^{14} \equiv 6(\bmod 19)$.

But this is remainder of only $\displaystyle 2^{2^{17}}$ if you want to find remainder of $\displaystyle 2^{2^{17}}+1$ just add 1.