if $(-1)^a5^b\equiv (-1)^c5^d \text{(mod $2^l$)},l\geq 3$, why is then $(-1)^a\equiv (-1)^c \text{(mod $4$)}$ (Question on Ireland&Rosen)?

Asked Feb 02 '25 at 16:05

Active Feb 02 '25 at 20:51

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This is from the book A Classical Introduction to Modern Number Theory by Ireland and Rosen. In theorem 2' (note the ') on page 43 of chapter 4, the authors state amidst a proof of the theorem that

If $(-1)^a5^b\equiv (-1)^c5^d \text{(mod $2^l$)},l\geq 3$, then $(-1)^a\equiv (-1)^c \text{(mod $4$)}$ implying that $a\equiv c \text{(mod $2$)}$.

Why is this true? How does one get to mod $4$ and then to mod $2$?

edited Feb 02 '25 at 20:51

Bill Dubuque

282,220

asked Feb 02 '25 at 16:05

Wasradin

1,619

For going from $\mod 4$ to $\mod 2$, a simple case by case ($a$ once odd, once even) suffices – Kraken Feb 02 '25 at 16:10
2

If $x\equiv y\pmod {2^l}$ then $x\equiv y\pmod 4.$ But $5^m\equiv1\pmod 4.$ – Thomas Andrews Feb 02 '25 at 16:13
@ThomasAndrews Ah, yes. Of course. It seems to take a while to train the eye to see -- or the brain to think -- the proper way when dealing with congruence structures. – Wasradin Feb 02 '25 at 16:16
$l\ge ,2\Rightarrow 4\mid 2^l,,$ so by 1st dupe: your congruence $!\bmod 2^l$ persists $!\bmod 4,,$ where $, 5^n\equiv 1^n\equiv 1,,$ so by congruence laws in 2nd dupe it's equivalent to $, (-1)^a\equiv (-1)^c,,$ so $,{\rm ord}_4(-1)=\color{#c00}{2}\Rightarrow$ $,a\equiv c\pmod{!\color{#c00}2},$ by mod order reduction Theorem (converse) in 3rd dupe. $\ \ $ – Bill Dubuque Feb 02 '25 at 20:48

if $(-1)^a5^b\equiv (-1)^c5^d \text{(mod $2^l$)},l\geq 3$, why is then $(-1)^a\equiv (-1)^c \text{(mod $4$)}$ (Question on Ireland&Rosen)?

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