Archive for July, 2013

Earlier today I met a person, well how do I describe him?  Perhaps the most smartest guy I ever met.  Who is he?  The famous Dr Koopa Koo.  I remembered I knew him from the uwants forum couple years ago, maybe around 2008?  Can’t remember exactly when.  Back then I only knew him as apook (his username on uwants) and knew he is a phenomenon in math.  We just knew each other through the web until today I finally have a chance to meet him in person in his lecture and have lunch with him after.  It’s a great honor to be able to sit in his class and I am sure I have learnt something already from him today.  And I will continue going to his lecture whenever possible because he is really that awesome!

During his lecture on number theory, all his audience are the gifted students from hong kong secondary school around the age of 12 to 18 I believed.  Although they are the “brains” in their school maybe with an IQ over 200, Koopa is able to make them feel like a human again in his class.  Even I felt stupid in his lecture.

Maybe I wasn’t a gifted person afterall but I would like to try his 5stars hard problem.  Here’s the problem if I remembered correctly (I have a very bad memory coz I am getting old) 🙂

Problem

Find all integer solution to

4y^3=3x^2+1

Solution

Of course one should be able to spot the trivial solution right away.

(1, 1) \longrightarrow 4(1)^3=3(1)^2+1

Now the question is can we find more solution other than the trivial one?

Let’s look at the left hand side, it is a multiple of 4, that means it is even.  This implies the right hand side also even and hence 3x^2 must be odd and therefore x must be odd as well.

So we let

x=2k+1, \forall k \in \mathbb{Z}

then,

3x^2+1=3(2k+1)^2+1=3(4k^2+4k+1)+1

=12k^2+12k+4=4y^3

3k^2+3k+1=y^3

Here is the tricky part, by the Binomial Theorem

(k+1)^3-k^3=y^3

(k+1)^3=y^3+k^3

By the Fermat’s Last Theorem (it’s proven in 1994), it states that there is no integer solution to

a^n+b^n=c^n, \quad \mbox{for } n > 2

\therefore k=0

\therefore y=1, x=\pm 1

Therefore (\pm 1, 1) is the only solution to this Diophantine equation. \square