# Math Help - Basic Question regarding Simplification of congruences

1. ## Basic Question regarding Simplification of congruences

Hi, I have just started learning the fundamentals of Number Theory and I'm having some trouble wrapping my head around a few steps that were presented in my textbook.

the first question contains a simplification of:

7x = 1(mod 8)

to

-x = 1(mod 8)

a side note states that this is possible because 7=-1(mod 8)

Now, if 7 were congruent to -1(mod 8), and since -1(mod 8) is congruent to -1 (this is what I gathered from their next step), would this imply that 7 is congruent to -1, and hence the next step simply replaces 7 with -1?

The second question contains the following step of simplification:

5(25^n)(mod 14)

to :

5(11^n)(mod 14)

I can see that the base has reduced by 14, but I am having trouble understanding why exactly 11^n would give the same remainder as 25^n when divided by 14.

I know that that a^n=b^n (mod 14), as well as the other "rules", but I just can't figure it out.

2. Originally Posted by Wandering
Hi, I have just started learning the fundamentals of Number Theory and I'm having some trouble wrapping my head around a few steps that were presented in my textbook.

the first question contains a simplification of:

7x = 1(mod 8)

to

-x = 1(mod)

a side note states that this is possible because 7=-1(mod 8)

Now, if 7 were congruent to -1(mod 8), and since -1(mod 8) is congruent to -1 (this is what I gathered from their next step), would this imply that 7 is congruent to -1, and hence the next step simply replaces 7 with -1?
you are right but one little mistake here is that you have said "...since -1(mod 8) is congruent to -1....". we never say that.
rather you say '-1 is congruent to -1 (mod 8)'.
you can replace 7 with -1 because of the elementary theorems of modular arithmetic. In full it will read:
$7 \equiv -1 (mod \, 8)$
and, $x \equiv x(mod \, 8)$
so, $7x \equiv -x (mod \, 8)$
do you see which theorem is used here?

Originally Posted by Wandering
The second question contains the following step of simplification:

5(25^n)(mod 14)

to :

5(11^n)(mod 14)

I can see that the base has reduced by 14, but I am having trouble understanding why exactly 11^n would give the same remainder as 25^n when divided by 14.

I know that that a^n=b^n (mod 14), as well as the other "rules", but I just can't figure it out.

$25 \equiv 11(mod \, 14)$
so, $25^n \equiv 11^n(mod \, 14)$(why?)
so, $5(25^n) \equiv 5(11^n) (mod \, 14)$(why?)