Why does p divides $2^{2(q-1)}-1$ if p divides $2^{q-1}+1$

Question

Why does p divides $2^{2(q-1)}-1$ if p divides $2^{q-1}+1$. The assumptions I have on $p$ and $q$ are that both are uneven and both are prime numbers. Also, besides having the proof, I'm tyring to have the "intuition" between this but I'm struggling.

Thanks!

The only thing that I found that could maybe be useful is the following: if p divides $2^{q-1}+1$ then p divides $2^{q-1}+2-1$ so p divides $2(2^{q-2}+1)-1$ — FluidMechanics Potential Flows, Feb 26 '22 at 11:21
By $,m=-1,$ in linked Factor Theorem: $, n+1\mid n^2-1.,$ OP is special case $,n = 2^{q-1}.\ $ — Bill Dubuque, Feb 26 '22 at 13:13
So $,p\mid n+1\mid n^2-1\Rightarrow p\mid n^2-1,$ by transitivity of divisibility (2nd linked dupe) — Bill Dubuque, Feb 26 '22 at 13:24

score 4 · Answer 1 · answered Feb 26 '22 at 11:39

4

Hint:$$2^{2(q-1)}-1=(2^{q-1}-1)(2^{q-1}+1)$$

answered Feb 26 '22 at 11:39

Shivam M

322

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 26 '22 at 13:13

score 1 · Accepted Answer · answered Feb 26 '22 at 11:48

1

$p\:|\:2^{q-1} + 1$ implies $2^{q-1} + 1 = ap$ for some integer $a$. Notice that $$2^{2(q-1)} - 1 = (2^{q-1} - 1)(2^{q-1} + 1) = (2^{q-1} - 1)ap.$$ Hence $2^{2(q-1)} - 1$ is an integer multiple of $p$, meaning that $p$ divides $2^{2(q-1)} - 1$.

answered Feb 26 '22 at 11:48

Ho Wing Yip

26

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 26 '22 at 13:13

Why does p divides $2^{2(q-1)}-1$ if p divides $2^{q-1}+1$

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