# Number Theory

• Feb 16th 2010, 05:03 PM
meshel88
Number Theory
Let n and a be positive integers and let p be a prime number. The p^a is said to exactly divide n. If p^a exactly divides n and p^a+1 does not exactly divide n. Assume p^a exactly divides m and p^b exactly divides n.

What power of p exactly divides m+n. Prove.
• Feb 16th 2010, 05:17 PM
icemanfan
Lemma: If $p^a$ exactly divides n, then no greater power of p divides n, and conversely, if $p^a | n$ and no greater power of p divides n, then $p^a$ exactly divides n.

The answer to your question is obviously a + b, but why? Can you use the lemma to show how?
• Feb 16th 2010, 05:36 PM
meshel88
yes you can use the lemma.
• Feb 16th 2010, 05:53 PM
icemanfan
It is clear that if $p^a | m$ and $p^b | n$, then $p^{(a+b)} | mn$. Suppose $p^{(a+b+k)} | mn$ for some positive integer k and $p^a, p^b$ both exactly divide m and n, respectively. Then since a is the greatest power of p that divides m, it must be true that $p^{(b+k)} | n$, which is impossible by the lemma. Therefore, $p^{(a+b)}$ divides mn exactly.

Edit: never mind, I had the question wrong.
• Feb 16th 2010, 06:05 PM
icemanfan
I don't think you have enough information.

Consider the following series of integers m, n:

$m = p(p - 1), n = p, m + n = p^2$

$m = p(p^2 - 1), n = p, m + n = p^3$

$m = p(p^3 - 1), n = p, m + n = p^4$

etc.

In each case it is clear that p exactly divides both m and n, but the power of p that exactly divides m + n is not determined by this fact.