Give an example to show that ## a^{k}\equiv b^{k}\pmod {n} ## ....

[Math100]

Homework Statement

Give an example to show that ## a^{k}\equiv b^{k}\pmod {n} ## and ## k\equiv j\pmod {n} ## need not imply that ## a^{j}\equiv b^{j}\pmod {n} ##.

Relevant Equations

None.

Disproof:

Here is a counterexample:
Let ## a=2, b=3, k=2, j=7 ## and ## n=5 ##.
Then ## 2^{2}\equiv 3^{2}\pmod {5} ## and ## 2\equiv 7\pmod {5} ##.
But note that ## 2^{7}\not\equiv 3^{7}\pmod {5} ##.
Thus ## a^{j}\not\equiv b^{j}\pmod {n} ##.
Therefore, ## a^{k}\equiv b^{k}\pmod {n} ## and ## k\equiv j\pmod {n} ## does not imply that ## a^{j}\equiv b^{j}\pmod {n} ##.

 

[anuttarasammyak]

Another counter example
Say a is any decimal positive integer that ends with 1 and b is any decimal positive integer that ends with 3.
[tex]a^4 \equiv b^4 \equiv 1 (mod \ 10)[/tex]
[tex]4 \equiv 14 (mod \ 10)[/tex]
[tex]a^{14} \equiv 1(mod \ 10), b^{14} \equiv 9 (mod \ 10)[/tex]

 

[SammyS]

Homework Statement:: Give an example to show that ## a^{k}\equiv b^{k}\pmod {n} ## and ## k\equiv j\pmod {n} ## need not imply that ## a^{j}\equiv b^{j}\pmod {n} ##.
Relevant Equations:: None.

Disproof:

Here is a counterexample:
Let ## a=2, b=3, k=2, j=7 ## and ## n=5 ##.
Then ## 2^{2}\equiv 3^{2}\pmod {5} ## and ## 2\equiv 7\pmod {5} ##.
But note that ## 2^{7}\not\equiv 3^{7}\pmod {5} ##.
Thus ## a^{j}\not\equiv b^{j}\pmod {n} ##.
Therefore, ## a^{k}\equiv b^{k}\pmod {n} ## and ## k\equiv j\pmod {n} ## does not imply that ## a^{j}\equiv b^{j}\pmod {n} ##.

Looks good.
You might want to at least demonstrate that you actually know that ## 2^{7}\not\equiv 3^{7}\pmod {5} ## .

##2^7 \equiv 3 \pmod 5 ## whereas ##3^7 \equiv 2 \pmod 5 ##

 

[fresh_42]

Homework Statement:: Give an example to show that ## a^{k}\equiv b^{k}\pmod {n} ## and ## k\equiv j\pmod {n} ## need not imply that ## a^{j}\equiv b^{j}\pmod {n} ##.
Relevant Equations:: None.

Disproof:

Here is a counterexample:
Let ## a=2, b=3, k=2, j=7 ## and ## n=5 ##.
Then ## 2^{2}\equiv 3^{2}\pmod {5} ## and ## 2\equiv 7\pmod {5} ##.
But note that ## 2^{7}\not\equiv 3^{7}\pmod {5} ##.
Thus ## a^{j}\not\equiv b^{j}\pmod {n} ##.
Therefore, ## a^{k}\equiv b^{k}\pmod {n} ## and ## k\equiv j\pmod {n} ## does not imply that ## a^{j}\equiv b^{j}\pmod {n} ##.

What @SammyS said, looks good. To know ##2^7=128\equiv 3\pmod 5## and ##3^7=2187\equiv 2\pmod 5## would have helped a lot. Most people have only the first, say four or five, powers in mind. It could be omitted in a book but should be told in an exercise.