It's been a long time...

by t0rajir0u, Mar 30, 2007, 3:56 PM

Let's talk about a useful technique for dealing with certain types of multiplicative number theory problems.

Definition: For a positive integer $n$, let $e_{p}(n)$ be the greatest $k$ such that $p^{k}| n$.

Corollary: $n = \prod_{p \in \mathbb{P}}p^{e_{p}(n)}$

Corollary of Equality: $a = b \Leftrightarrow e_{p}(a) = e_{p}(b) \forall p \in \mathbb{P}$

Corollary: $e_{p}(ab) = e_{p}(a)+e_{p}(b)$

With this very simple tool we can already prove some useful identities.

Lemma 1: $e_{p}( gcd(a, b) ) = min(e_{p}(a), e_{p}(b))$

Lemma 2: $e_{p}( lcm(a, b) ) = max( e_{p}(a), e_{p}(b))$

Lemma 3: $min(x, y)+max(x, y) = x+y$ (simply assume WLOG $x \le y$)

Lemma 4: $gcd(a, b) lcm(a, b) = ab$

Proof: Using the corollary of equality, this reduces to Lemma 3.

Lemma 5: Let $S = \sum_{i=1}^{k}a_{i}$. We have the series of identities

$min( a_{i})+max(S-a_{i}) = S$
$min( a_{i}+a_{j})+max(S-a_{i}-a_{j}) = S$
$min( a_{i}+a_{j}+a_{k})+max(S-a_{i}-a_{j}-a_{k}) = S$
...
$min( S-a_{i})+max( a_{i}) = S$

Again, these are easily proven by assuming WLOG $a_{1}\le a_{2}\le ... \le a_{k}$.

Lemma 6: Let $\{ x_{i}\}$ be a set of positive integers, and let $P = \prod_{i=1}^{k}x_{i}$. Then by the corollary of equality we have the identities

$gcd(x_{i}) lcm \left( \frac{P}{x_{i}}\right) = P$
$gcd(x_{i}x_{j}) lcm \left( \frac{P}{x_{i}x_{j}}\right) = P$
...
$gcd \left( \frac{P}{x_{i}}\right) lcm( x_{i}) = P$

These identities are very nice. For $k = 3$ they become

$gcd(a, b, c) lcm(ab, bc, ca) = abc$
$gcd(ab, bc, ca) lcm(a, b, c) = abc$

And for $k = 4$ they are

$gcd(a, b, c, d) lcm(abc, bcd, cda, dac) = abcd$
$gcd(ab, ac, ad, bc, bd, cd) lcm(ab, ac, ad, bc, bd, cd) = abcd$
$gcd(abc, bcd, cda, dab) lcm(a, b, c, d) = abcd$

Note the relation to binomial coefficients. The missing coefficients of $1$ give trivial identities such as $gcd(1) lcm(abcd) = abcd$.


Of course, $e_{p}(n)$ has other uses. Often in problems involving divisibility and/or factorials it is more convenient to consider an equivalent problem given by the corollary of equality than the original problem.

Problem: Prove that $\frac{ \left( \sum a_{i}\right)! }{ \prod (a_{i}!) }$ is an integer.

Solution: Using the corollary of equality, this reduces to showing that

$\sum e_{p}( a_{i}! ) \le e_{p}(( \sum a_{i})! )$

But if you know that

$e_{p}( k! ) = \sum \lfloor \frac{k}{p^{i}}\rfloor$

Then all it takes is another lemma.

Lemma: $\left\lfloor \frac{ \sum x_{i}}{d}\right\rfloor \ge \sum \left\lfloor \frac{x_{i}}{d}\right\rfloor$


Practice Problem 1: If $m, n$ are positive integers with $m$ odd, determine $gcd(2^{m}-1, 2^{n}+1)$.

Practice Problem 2: Prove that

$P_{n, r}(x) = \frac{ (1-x^{n+1})(1-x^{n+2})...(1-x^{n+r}) }{(1-x)(1-x^{2})...(1-x^{r})}$

Is a polynomial in $x$ of degree $nr$. (Hint: What "primes" are applicable here?)

Comment

3 Comments

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:)
I have a problem for you.
IF ab+1/ a^2 + b^2 , prove that

(a^2+b^2)/ab+1 is a perfect square.

by rohit, May 7, 2007, 5:48 AM

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It's an very old and famous problem in IMO(1976).Killing it by using Vieta's jumping is a wondeful method! Can you prove the general problem? :wink:

by Anonymous, Dec 11, 2007, 11:26 PM

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Hey,
Can't we show there is a combinatorial interpretation for this:$ \frac{ \left( \sum a_{i}\right)! }{ \prod (a_{i}!) }$ and hence must be an integer.

by isomorphism, Dec 30, 2007, 8:26 AM

Various mathematical thoughts, ranging from problem-solving techniques to attempts to tie together related concepts.

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