Let . Show that is a multiple of for .
How to do this question..?
, so the solution follows at once from the following
Lemma: For every , we have that , for some
From the above, it turns out that the powers of p cancel each others in pairs (one from the numerator, one from the denominator), and doesn't cancel COMPLETELY with no power of from the denominator EXCEPT, perhaps, with ...as , it follows that and then and we're done.
Tonio