different ways of calculating mod values

Question

Calculate $$10^{(10^{10})} \pmod 7$$

With Fermat:

$$10^6 \equiv 1 \pmod 7 \Rightarrow 10^{6t} \equiv 1 \pmod 7$$

$$ \begin{eqnarray*} 10 & \equiv & -4 \pmod6\\ 10^{10} & \equiv & (-4)^{10}=[(4)^2]^5=(-2)^5=-32 \equiv 4 \pmod6\\ 10^{10} & = &6t+4 \\ 10^{{10}^{10}} & = & 10^{(6t+4)} \equiv 10^4 \equiv (-3)^4=81 \equiv 4 \pmod7\\ \end{eqnarray*}$$

Why did he write $10 \equiv -4 \; mod \; 6$, which lead to $10^4 \equiv (-3)^4 \pmod 6$ instead of positive $3^4$? Is what he wrote correct? Doesn't 10= 4 mod(6)? and so on.

As well,

Does $mod \; n$ always repeat when the value reaches 1? if so, why?

Answer 1

If 10 ≡ -4 (mod 6) then 10 + 4 ≡ 0 (mod 6), but this is false since 14 is not divisible by 6. So 10 ≡ -4 (mod 6) is false. On the other hand, 10 ≡ 4 (mod 6) is true since 10 - 4 is divisible by 6.

For the second question, if $a^b$ ≡ 1 (mod n) for some b, then $a^{b+k}$ ≡ $1 \times a^k ≡ a^k$ (mod n). So yep, it does indeed "repeat". In fact, $b$ in this case is called the "exponent of the multiplicative group of integers modulo n". The smallest value of $b$ can be calculated using the Carmichael function (if $a$ and $n$ are coprime). See: http://en.wikipedia.org/wiki/Carmichael_function

Answer 2

Yes, there is probably a mixup of -2 and 4 as possible remainders of 10 mod 6. In the end it does not matter, since the 4 gets squared.

The periodicity is a consequence of the finiteness of the multiplicative group, it is formalized in the little theorem of Fermat.