Note 1
[N] Miscellaneous Goodies
This note discusses some of the miscellaneous Number Theory topics that are not necessarily the major part in olympiad NT, but would be very useful when solving problems. At least for KMO, I am aware that it is generally required to prove many of the contents here during the write-ups, but some problems are virtually impossible to solve without previous knowledge on these topics!
1.1 \(p\)-adic Valuation & LTE Lemma
LTE Lemma, which stands for Lifting the Exponent Lemma, concerns about \(p\)-adic valuation of two integers. Before we get into the lemma, let’s discuss about \(p\)-adic valuation.
Definition 1.1.1
The \(\mathbf {p}\)-adic valuation of an integer \(n\) and prime \(p\), denoted as \(\nu _p (n)\), is the largest possible integer \(a\) such that \(p^a \mid n\).
Some examples could be \(\nu _2 (48) = 3\) and \(\nu _3 (54) = 3\). It is worth noting that some sources write this as \(V_p (n)\), so they most likely refer to the same thing.
A very famous example would be Legendre’s formula which states the following equation for positive \(n\) and prime \(p\). \[ \nu _p (n!) = \sum _{k=1}^\infty \left \lfloor \frac {n}{p^k} \right \rfloor \]
This is actually the same as the classic competition style problems that asks you the number of \(p\) in \(n!\). I will skip the proof because I think it is pretty self-evident.
Another very famous application is understanding why \(\binom {n}{r}\) is an integer. Obviously combinatoric-wise, its definition makes it an integer, but algebraic-wise, it is pretty interesting considering the fact that \(\binom {n}{r} = \frac {n!}{r! (n-r)!}\). To show this, it suffices to prove the following inequality for any prime \(p\). \[ \nu _p (n!) \geq \nu _p (r!) + \nu _p ((n-r)!) \] By Legendre’s formula, the inequality converts as \[ \sum _{k=1}^\infty \left \lfloor \frac {n}{p^k} \right \rfloor \geq \sum _{k=1}^\infty \left ( \left \lfloor \frac {r}{p^k} \right \rfloor + \left \lfloor \frac {n-r}{p^k} \right \rfloor \right ) \] and this is true because \(\lfloor a + b \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor \). With this notation and concept in mind, lets get into LTE Lemma!