For any odd prime $p \gt 1$, show that $2^p \equiv 8 \pmod {12}$

Question

I tried simplifying the problem to $2^p \equiv 2 \pmod 3$ then to $2^{p-1} \equiv 1 \pmod 3$. But I am unable to see how to go further or which property I could use. Any help on how to move ahead is appreciated. Thanks!

Answer 1

By Chinese remainder theorem, $2^p\equiv 8\ ($mod $12)$ is equivalent to $$2^p\equiv 0\ (\text{mod }4) \text{ and }2^p\equiv 2\ (\text{mod }3).$$ If $p$ is an odd integer at least 3, the former equation is clear and so is the latter since $$2^p=4^n\cdot 2\equiv 2\ (\text{mod }3)$$ if $p=2n+1$. Therefore the assertion holds.

Answer 2

It's true for $p=3$ so assume $p>3$, then $$2^p\equiv 8\equiv 2^3\pmod{ 12}\iff 2^{p-2}\equiv2\pmod 3$$ and since $(2,3)=1$, we only need to prove $$ 2^{p-3}\equiv1\pmod 3$$ But $p$ is odd (and we only need the its oddness) so $p-3$ is even. Therefore $$2^{p-3}\equiv(-1)^{p-3}\equiv 1\pmod 3.$$

Answer 3

By examination, $2^3=8\equiv 8 \bmod 12$

All other odd primes have the form $6k+1$ or $6k+5$

Important fact: $4$ is an idempotent $\bmod 12$. That means that $4$ raised to any power is $\equiv 4 \bmod 12$. To prove this it is only necessary to show that $4^2=16\equiv 4 \bmod 12$. Thereafter multiplying that product repeatedly by $4$ (i.e. raising $4$ to a higher exponent) changes nothing.

Case 1: $2^{6k+1}=2\cdot 64^k \equiv 2\cdot 4^k \equiv 2\cdot 4 \equiv 8 \bmod 12$

Case 2: $2^{6k+5}=32\cdot 64^k \equiv 8\cdot 4^k \equiv 8\cdot 4 \equiv 32 \equiv 8 \bmod 12$

Thus for all odd primes $p$, $2^p \equiv 8 \bmod 12$ QED

Note that for numbers of the form $6k+1$ or $6k+5$, it is not necessary that they be prime for this result to hold.

Answer 4

The congruence $2^p\equiv 8\bmod 12$ means that $12$ divides $2^p-8$. After factoring out a factor $4$ the latter divisibility is equivalent to $3$ divides $2^{p-2}-2$ which in turn can be rewritten as the congruence $$ 2^{p-2}\equiv2\bmod 3. $$ But $2\equiv-1\bmod 3$ so we can rewrite the congruence as $$ (-1)^{p-2}\equiv-1\bmod 3 $$ which is obviously true since $p-2$ is odd.