Two little problems (ii)

Problem 2. Find the solutions of $x^6 + y^6 + z^6 = 6xyz - 3$.

Equalities like this ($xyz$? Sixth powers? I don't know the formula for that!) can look intimidating, but a cool head and some experience will make dealing with such unfamiliar things a bit more familiar.

Tempted to simplify the right hand side, we divide by three:
\[
\frac{x^6 + y^6 + z^6}{3} = 2xyz - 1
\]
The key of the exercise is to notice how the left hand side can be seen as an arithmetic average of 3 values, $x^6$, $y^6$ and $z^6$. Given how either the arithmetic average is always greater than the geometric average, or all the values are equal, we can consider the inequality:
\[
\sqrt[3]{x^6y^6z^6} \leq \frac{x^6 + y^6 + z^6}{3}
\]
And we simplify:
\[
x^2y^2z^2 \leq \frac{x^6 + y^6 + z^6}{3}
\]
We now define $xyz = t$, and proceed to exchange the fraction by the equivalence we first saw:
\[
t^2 \leq 2t - 1
\]
Therefore $(t-1)^2 \leq 0$, which is only possible for $t=1$ (since any other value will result in a positive square).

Now, since the product of the three variables is 1, $\sqrt[3]{x^6y^6z^6} = \sqrt[3]{1^6} = 1$, and going back to the initial equality, $2xyz - 1 = 2 - 1 = 1$, which is equal as well to the arithmetic average, we find that both the $AM$ and the $GM$ are equal, and therefore their elements are equal as well. Then:
\[
x^6 + x^6 + x^6 = 3x^6 = 3
\]
And so $x^6 = 1$. We find six solutions for $x$, all of them being either 1 or -1, but we have to restrict to combinations such that $xyz = 1$, which is left as a petty exercise for the reader.

We did this problem in class, and everyone was quite fascinated with how the problem simply unraveled with the use of such elementary methods. Admittedly, it's a very synthetic problem, but it testes your ability to see that which is not immediately obvious to the eye.

Problem 3. Any power $n^k$ can be expressed as the sum of $n$ consecutive odd integers.

This one comes from proving the $2^k$ case, and then being asked to do the same with $3^k$. It seemed too good to be true, but as it turns out, the proof is quite simple.

We assert the theorem:
\[
n^k = (2q + 1) + (2q + 3) + (2q + 5) + ... + (2q + (2n - 1))
\]
Let's reorder this:
\[
n^k = n2q + (1 + 3 + 5 + 7 + ... (2n - 1))
\]
It is a fact that the sum of the first $r$ odds is equal to $r^2$ (it's a one-line proof), so:
\[
n^k = n2q + n^2
\]
We find $q$:
\[
q = \frac{n^{k-1} - n}{2}
\]
And then verify the identity:
\[
n^k = n2\left(\frac{n^{k-1} - n}{2}\right) + n^2 = n(n^{k-1} - n) = n^n - n^2 + n^2 = n^k
\]
And we're done; the identity derived from the sum of $n$ consecutive odd integers is proved to be equal to $n^k$!

Snow

Today has been a snowy day (not in the most rigorous sense of the word) all around the country. The camera struggled with such little amount of light, but it pulled this one off somehow, which looks cool enough.

Visit to the MNAC (National Art Museum of Catalonia)

This weekend i visited the National Art Museum of Catalonia for a few hours. Sadly, i couldn't take the camera with me, so all pictures were taken with my phone.

[algebra] A little problem (ii)

I'm following the great course on abstract algebra offered online by Harvard, lectured by Benedict Gross. So far, i'm having a great time. I'm trying to do the exercises (Algebra, Artin, 1st ed. 1991), to gain a better understanding and intuition on the matter, so i'll be posting exercises that i find interesting. Mind you, i'm just getting started, so interesting may be a bit too far-fetched...

Exercise (2.2.20.a): Let $G$ be an abelian group, and $a,b \in G$. Let $m,n$ be the orders of the groups generated by $\langle a \rangle$ and $\langle b \rangle$. What can you say about the order of $\langle ab \rangle$?

We consider the elements generated by $\langle ab \rangle$:

• 1 (identity).
• $ab$.
• $(ab)^2 = abab$. Since the group is commutative, we can reorder like this: $aabb = a^2b^2$.
• $(ab)^k = a^kb^k$, more generally.

Therefore, once we reach either $(ab)^m$ or $(ab)^n$, whichever's the smallest of $m$ and $n$, we will be left with the rest of the cyclic group generated by $a$ or $b$. So, we can conclude that the order of $\langle ab \rangle$ is in fact the order of $a$ or $b$.

A little problem (i)

Problem 1. Demostra que ${1992 \choose 1492}$ no és divisible per 500 -- Prove that ${1992 \choose 1492}$ is not divisible by 500.

Given that we're asked to prove divisibility, decomposing the numbers in prime factors is a nice idea in order to get somewhere. $1992 = 2^3 \cdot 3 \cdot 83$, $500 = 2^2 \cdot 5^3$ and $1492= 2^2 \cdot 373$. Let's now analyze the binomial expansion:
\[\frac{1992!}{(1992-1492)! \cdot 1492!} = \frac{1992!}{500! \cdot 1492!}\]
We're asked to prove that the quotient will not be divisible by 500, and so after we simplify the expression, we must find that there are less than two 2s and three 5s left. Proving one of these two conditions will solve the problem. However the denominator simplifies with the numerator, we should find that there are not enough factors left to form the number 500.

In this case, intuition told me that there were going to be enough 2s left, and so i jumped straight into finding out how many 5s were going to be left. Either way, the process is similar.

We must count the amount of 5s we will have in the numerator, in that gigantic factorial. To do so, we have to check how many multiples of $5$, $5^2$, $5^3$ ... there are (until we get to the point where $5^x$ is more than 1992). We can take the fifth root to find $x$, like so:
\[ \lfloor \sqrt[5]{1992} \rfloor = 4 \]
So, we have to count the multiples of up to $5^4$:
\[ \sum_{n=1}^{4} \lfloor \frac{1992}{5^n} \rfloor = 398 + 79 + 15 + 3 = 495  \]
Therefore, we have a total of $495$ 5s somewhere in the numerator. Fairly enough, we simply have to check how many 5s we have in the denominator and see if the difference is less than 3. For 500, we take the fifth root and find that we can find multiples of up to $5^3$. We take the fifth root again for 1492 and 4 is the exponent.
\[ \sum_{n=1}^{3} \lfloor \frac{500}{5^n} \rfloor = 100 + 20 + 4 = 124 \]
\[ \sum_{n=1}^{4} \lfloor \frac{1492}{5^n} \rfloor = 298 + 59 + 11 + 2 = 370 \]

And nicely enough, the total amount of 5s we have in the denominator is 494, which leaves us with a single, poor 5 once the fraction is simplified. Not enough to form a 500, and so the problem is solved.

Easy (relatively, but uncommon, surely) problems like these help keep the brain alive in the cold winter school days. Take a stand and keep yours active! This is a problem from the Catalonian regional phase of the math Olympiads, back in 1992.

Some pictures

Not much time left until the restart of routine, so i'm already psychologically torturing myself over it, i can't help it.

A night in the summer

A fantastic panoramic of the Milky Way, taken one night this summer, during a stay in the mountains. All credits to the great astrophotographer Juan Carlos Casado.