Divisibility

hope this is cent percent rite

(aλ + δ )^k+1 = a^k.β

for k=1 a|δ²

for k=2 a²|δ³
( as (aλ+δ)³ /a² = aμ + ρ +ξδ²/a + δ³/a² )

similarly for any k

a^k| δ^k+1

now as δ^k+1 = β. a^k and δ<a

for k→∞ , β=0 → δ=0

hence proved

See β→0, but a^k→∞. So what tells you that δ=0?

β→0 implies β=0 as β is integer

im sure this time its flawless :P

that 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 ord_r(p)

Now, if for some prime p, we have ord_p(a)>ord_p(b)

Let ord_p(a) = m and ord_p(b) = n

Now for all k>n/(m-n) we will have ord_p(a^k) > ord_p(b^k+1)

which means a^k does not divide b^k+1 for k>n/(m-n)

why the (phew!) ?