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Suppose we know that $$a \equiv b (mod \ k)$$, what must be true for $$a \equiv b (mod \ j)$$

So far this is what I have:

We know that these two equivalences can be rewritten as :

$$k |(a-b)$$ and $$j | (a-b)$$

Thus we can say that $$kn = (a-b)$$ for some integer n

and $$jp = (a-b)$$ for some integer p.

I set them equal to each other:

$$ kn = jp$$

I'm not even sure if this is the right approach. Could someone help me out?

Thanks in advance.

Pandamonium
  • 975
  • If $a$ is arbitrary (while subjecting to $a\equiv b(\mod k)$, then I don't think you can condition $n$. In particular, $a$ can just be $b+k$, so implies $j|k$. – Kim Jong Un Sep 11 '14 at 04:11
  • i guess your right , $k$ and $j$ are just some factors of $(a-b)$ – avz2611 Sep 11 '14 at 04:13
  • It's fine if $j$ is a divisor of $k$. You can always replace the modulus of a congruence relation by a smaller number dividing it. – Zavosh Sep 11 '14 at 04:36
  • @AlexWong Beats me, ie answer looks like "nothing". Are you sure there is no other information given? – almagest Sep 11 '14 at 04:43

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