bacasauvfl

2022-04-28

Let x be an integer and let d be a factor of $2{x}^{2}-1$. Prove that ${d}^{2}\equiv 1\mathrm{mod}16$

hoppledhsy

Beginner2022-04-29Added 13 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 $2{x}^{2}-1\iff 2$ is a quadratic residue modulo p.

$\iff p\equiv \pm 1\mathrm{mod}8$. Obviously 2 is not a factor of $2{x}^{2}-1$.

So d is a factor of $2{x}^{2}-1\iff d\equiv \pm 1\mathrm{mod}8\iff {d}^{2}\equiv 1\mathrm{mod}16$

The last equivalence follows because if $8\mid d\pm 1$ then $16\mid (d+1)(d-1)={d}^{2}-1$.

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