Congruent Integers

  • Thread starter scottstapp
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  • #1
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Homework Statement


I need to prove the following but have no idea how to do so.

Let a,b, k be integers with k positive. If a is congruent to b(mod n), then ak is congruent to bk (mod n).


Homework Equations


The hint given is that I can assume the following proposition is true and that I am supposed to use it to show the statement holds for k=2,3...

Proposition:
If a is congruent to b(mod n) and c is congruent to d(mod n) then a+c is congruent to b+d(mod n)

Thanks for your help, I am pretty lost on this so anything helps.
 

Answers and Replies

  • #2
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The directions of this problem seem to be pointing you to doing it by induction on k. For k = 1, the proposition is obviously true.

Assume the proposition is true for k = m. IOW, assume that
am [itex] \equiv [/itex] bm mod n.

Now use this assumption to show that the proposition is true for k = m + 1. I.e., that
am + 1 [itex] \equiv [/itex] bm + 1 mod n.

Something that might be helpful is that if p [itex] \equiv [/itex] q mod n, then p - q [itex] \equiv [/itex] 0 mod n.
 
  • #3
HallsofIvy
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Simpler, I think. [itex]a^k- b^k= (a- b)(a^{k-1}+ a^{k-2}b+ \cdot\cdot\cdot ab^{k-2}+ b^{k-1})[/itex]. Now use the fact that, since a= b (mod n), a- b is a multiple of n.
 
  • #4
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7,137
That's similar to the direction I was sending the OP, but didn't want to get dinged again for giving too much help...
 

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