Suppose that $ \large p $ is a prime, then
$$
\large 2^{p-1}\left(2^p -1 \right)
$$
ends with a 6 or 8.

Question

Suppose that $ \large p $ is a prime, then
$$
\large 2^{p-1}\left(2^p -1       \right)
$$
ends with a 6 or 8.

anonymous · Answer

let's consider $2^{p-1}(2^p-1)\mod10$... there is a clear pattern $2^n\mod 10$:$$2^0\equiv1\mod10\2^1\equiv2\mod10\2^2\equiv4\mod10\2^3\equiv8\mod10\2^4\equiv6\mod10\2^5\equiv2\mod10\\dots$$so clearly for odd $n$ we see that $2^n$ cycles $2,8,2,8,2\dots$. either $2^p\equiv2$ or $2^p\equiv8$

anonymous · Answer

when $p = 2$, its clearly true. we only have to prove it for odd primes. and we can use fermat theorem i guess ?

anonymous · Answer

if $2^p\equiv2$ then either $2^{p-1}\equiv1$ (for $p=1$) or $2^{p-1}\equiv6$. since $p$ is prime we know then that $2^{p-1}\equiv6$ ergo $2^p-1\equiv1$ and $2^{p-1}(2^p-1)\equiv6\cdot1\equiv6$

anonymous · Answer

if $2^p\equiv8$ then we know $2^{p-1}\equiv4$ ergo $2^p-1\equiv7$ so $2^{p-1}(2^p-1)\equiv4\cdot7\equiv8$ Q.E.D.

anonymous · Answer

yep, that'll do it :-)

anonymous · Answer

i understood $2^p \equiv 8$ case comletely
however i dont get the $2^p\equiv 2 $ case yet :|

anonymous · Answer

when $2^p\equiv2  \mod  10$,
$2^{p-1} \equiv 1  \mod  10$ ... how can we divide like this ? 10 is not a prime right

anonymous · Answer

I don't divide (clearly $2$ is non-invertible) but I just look at the pattern

anonymous · Answer

oh your first replyy...

anonymous · Answer

you can show very easily that $2^n$ is periodic for $n>0$, the only "odd" term is \(n=0\

anonymous · Answer

but yes

anonymous · Answer

so the whole proof is based on cyclicity... 
we look at what comes before the powers for remainders 2 and 8. got it :) and it looks great, thank you :)

anonymous · Answer

NP and yeah it's all about how the powers are periodic

anonymous · Answer

Here is a prrof that I know:
If p=2, then $ 2^{p-1} \left( 2^p-1    \right)=6$

if p>2, then
p=4m +1 or p= 4m +3  othereise, p is divisible by 2

Let us consider the first case p=4m +1

$$

2^{p-1} \left( 2^p-1    \right)= 2^{4m} \left( 2^{4m+1}-1    \right)=\

= 16^m \left( 2\, 16^m-1    \right)

$$
You can show by induction that $ 16^m $ always ends up in 6. so
$$

\left( 2\, 16^m-1    \right)

$$
ends up 1, So we are done in this case.
If p=4m +3

$$

2^{p-1} \left( 2^p-1    \right)= 2^{4m+2} \left( 2^{4m+3}-1    \right)=\

= 4\,16^m \left( 8 \, 16^m-1    \right)

$$

Now $ 4 \, 16^m $ ends in 4.
$ 8 \, 16^m -1 $ ends up in 7
So our number ends up in 8

anonymous · Answer

yuck! that looks a lot harder. you can make my part about periodicity rigorous very easily... and it's far less a hassle... :-p