Congruence of given statements proof

[lookingforhelp]

Assuming a,b,c,d, and n are integers with n>1 and that a is congruent to c (mod n), and b is congruent to d (mod n); prove that a^m is congruent to c^m ( mod n) for integers m>=1.
I think that I need to use induction for this proof, and the relation the relation that a+-b is congruent to c+-d (mod n) but I'm not sure if that is correct and how to do it. Thank you for the help.

 

[daon2]

Assuming a,b,c,d, and n are integers with n>1 and that a is congruent to c (mod n), and b is congruent to d (mod n); prove that a^m is congruent to c^m ( mod n) for integers m>=1.
I think that I need to use induction for this proof, and the relation the relation that a+-b is congruent to c+-d (mod n) but I'm not sure if that is correct and how to do it. Thank you for the help.

what purpose to b and d have??

Here is what I would prove, by induction if necessary,

\(\displaystyle a\equiv c \text{ (mod n)} \iff n|(a-c)\)

But \(\displaystyle (a-c) | (a^m-c^m)\) so by transitivity \(\displaystyle n | (a^m-c^m)\)

 

[HallsofIvy]

"b is congruent to d (mod n)" may be used in another problem, but, as daon2 said, there are no "b" or "d" in this problem. And no "induction" is needed, just that \(\displaystyle (a- b)(a^{m-1}+ a^{m-2}b+ \cdot\cdot\cdot+ ab^{m-2}+ b^{m-1})\).

 

[daon2]

"b is congruent to d (mod n)" may be used in another problem, but, as daon2 said, there are no "b" or "d" in this problem. And no "induction" is needed, just that \(\displaystyle (a- b)(a^{m-1}+ a^{m-2}b+ \cdot\cdot\cdot+ ab^{m-2}+ b^{m-1})\).

I was once a TA for an introductory proofs class, and the statement you wrote above was an exercise to be proved by induction. So it's YMMV for the assumption of that truth