341how do you get a^k|\delta^{k+1}?
If a and b are positive integers such that for all positive integers k, ak| bk+1, then prove that a|b
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11 Answers
9hope this is cent percent rite
(aλ + δ )k+1 = ak.β
for k=1 a|δ2
for k=2 a2|δ3
( as (aλ+δ)3 /a2 = aμ + Ï +ξδ2/a + δ3/a2 )
similarly for any k
ak| δk+1
now as δk+1 = β. ak and δ<a
for k→∞ , β=0 → δ=0
hence proved
341somethings need to be cleared up. β is not a constant. It is a function of k. So as k→∞, β(k)→0.
So, now how do we conclude that δ = 0?
9β→0 implies β=0 as β is integer
im sure this time its flawless :P
341that looks alright(phew!)
there another way - when we say p|q what we are saying is that if r is a prime that occurs in p, it occurs to a higher power in q. This power is denoted by ordr(p)
Now, if for some prime p, we have ordp(a)>ordp(b)
Let ordp(a) = m and ordp(b) = n
Now for all k>n/(m-n) we will have ordp(ak) > ordp(bk+1)
which means ak does not divide bk+1 for k>n/(m-n)