Show that $(-1)^a5^b \equiv (-1)^{a'}5^{b'} \mod 2^l$, $ l\geq 3$ implies $(-1)^a \equiv (-1)^{a'} \mod 4$ and $a\equiv a' \mod 2$

Asked Jun 21 '23 at 20:45

Active Jun 22 '23 at 05:07

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I am reading Ireland's A classical introduction to modern number theory. I find it hard to understand a part of a proof in the book. Can anyone explain it for me?

If $(-1)^a5^b \equiv (-1)^{a'}5^{b'}\mod 2^l$, $ l\geq 3$,$\,$ then $(-1)^a \equiv (-1)^{a'} \mod 4$, implying that $a\equiv a' \mod 2$.

edited Jun 22 '23 at 05:05

Bill Dubuque

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asked Jun 21 '23 at 20:45

zyy

By the first dupe and $4\mid 2^l,,$ the congruence persists $!\bmod 4,$ yielding $, (-1)^a\equiv (-1)^{a'}\pmod{!4},,$ so $,a\equiv a'\pmod{!\color{#c00}2},$ by $,{\rm ord}_{:!4}(-1)=\color{#c00}2,$ and mod order reduction (2nd dupe). $\ \ $ – Bill Dubuque Jun 22 '23 at 05:14

Show that $(-1)^a5^b \equiv (-1)^{a'}5^{b'} \mod 2^l$, $ l\geq 3$ implies $(-1)^a \equiv (-1)^{a'} \mod 4$ and $a\equiv a' \mod 2$

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