Find all integers $n$ such that $7 \mid (5^n -1).$

Question

Find all integers $n$ such that $7 \mid (5^n +1).$

From flt, $5^6\equiv 1 \pmod 7.$ but how to proceed from here?

You only have to find the smallest positive integer $\ k\ $ with $\ 5^k\equiv 1\mod 7\ $, which is $\ 6\ $. The solutions are then exactly the multiples of $\ k\ $, in this case the multiples of $\ 6\ $. This $\ k\ $ is also called the order of $\ 5\ $ modulo $\ 7\ $ , and is denoted by $\ k=ord_5(7)\ $ — Peter, Mar 12 '20 at 09:36
the question says (5^n + 1). I have understood for minus. But how Should I proceed with a plus sign? — Diwakar Naidu Gunti, Mar 12 '20 at 09:46

Mathsmerizing · Accepted Answer · 2020-03-12T09:52:16.057

0

I: For minus sign

$5^n-1={5^3}^{(\frac{n}{3})}-1$

$={(126-1)}^{(\frac{n}{3})}-1$

$={(7*18-1)}^{(\frac{n}{3})}-1$

$={(-1)}^{(\frac{n}{3})}-1$

For it to be zero, $\frac{n}{3}$=even i.e. 2k

so n should be of the form 6k

II: With a plus sign

$5^n+1={5^3}^{(\frac{n}{3})}+1$

$={(126-1)}^{(\frac{n}{3})}+1$

$={(7*18-1)}^{(\frac{n}{3})}+1$

$={(-1)}^{(\frac{n}{3})}+1$

With a plus sign for it to be zero $\frac{n}{3}$= odd i.e. 2k-1

so n should be of the form n=3(2k-1)

edited Mar 12 '20 at 09:52

answered Mar 12 '20 at 09:49

Mathsmerizing

thanks. But the question is 5^n +1. can you please solve for the plus sign? – Diwakar Naidu Gunti Mar 12 '20 at 09:52
@Diwakar Changed it to include plus also – Mathsmerizing Mar 12 '20 at 09:56

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