Check the solution to "Let x be an integer and let d..."

Dominique Pace

Answered question

2022-03-21

Let x be an integer and let d be a factor of

2 x^{2} - 1

. Prove that

d^{2} \equiv 1 b mod 16

lernarfnincln6g

Beginner2022-03-22Added 14 answers

If d is a factor of

2 x^{2} - 1

, then any ' factor p of d is a factor of

2 x^{2} - 1

.
Prime

p > 2

is a factor of

2 x^{2} - 1 \Leftrightarrow 2

is a quadratic residue modulo p.

\Leftrightarrow p \equiv \pm 1 b mod 8

. Obviously 2 is not a factor of

2 x^{2} - 1

.
So d is a factor of

2 x^{2} - 1 \Leftrightarrow d \equiv \pm 1 b mod 8 \Leftrightarrow d^{2} \equiv 1 b mod 16

The last equivalence follows because if

8 ∣ d \pm 1

then

16 ∣ (d + 1) (d - 1) = d^{2} - 1

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?