What's up people? As you can see, this isn't an applied maths post, as I promised before. I still have such a post in mind; basically, my idea is to create a system of differential equations that models all sorts of relationships based on the Sociotype. The idea is there, and so are the mathematical methods, I just need a bit of field work. But asking random people on the street to take an obscure test on the internet alongside their partners and then trying to quantify their enthusiasm and level of involvement in the relationship is a bit too much, so I'll probably end up posting the solution of the Gompertz equation or the prey-predator model.
But, until then, I decided I'd take a look at a subject I didn't really care about throughout high school: number theory. I devoted most of my time to algebra, and a bit of it to analysis, but, after seeing the beauty of these results, I really regret it. I'm probably gonna study some algebraic number theory in my free time to compensate for my lack of knowledge.
So there I was, in Bulgaria, surfing AoPS, when I saw some Number Theory problem, with a guy commenting that it's similar to Lagrange's 4 squares theorem. I checked it out on Wikipedia (great place for that, right?), and, apparently, it is solvable with quaternions, numbers of the form \(x=a+ib+jc+kd\), with \(i^2=j^2=k^2=ijk=-1\), and \(ij=k, jk=i, ki=j, ji=-k, kj=-i, ik=-j\).
The theorem states that any natural number can be written as a sum of 4 squares. More intuitively, it states that, if you have a sphere in 4 dimensions, then that sphere contains an integer point on its contour. In order to prove this, we shall use a sub-field of the quaternions, the Hurwitz quaternions, which are numbers of the form \(x=a+ib+jc+kd\), with \(a,b,c,d\) all integers or halves of odd numbers.
First, let us note that, if two numbers can be written as the sum of 4 squares, then their product can be written in that form, too. Indeed, presume that \(x=a^2+b^2+c^2+d^2, y=e^2+f^2+g^2+h^2\). The sum of 4 squares is the modulus of a quaternion with integer coefficients, and, thus, \(xy\) would be the modulus of the product of quaternions with integer coefficients, which, in turn, is the modulus of a quaternion with integer coefficients. This concludes the fact that \(xy\) can be written as a sum of 4 squares.
With this fact, we can resume to proving that every prime can be written as a sum of 4 squares. We know that any odd prime divides a number of the form \(1+l^2+m^2\). We shall now discuss the Hurwitz primes, which act literally like the normal prime numbers, but belong to the quaternions. They are literally prime quaternions.
Let us presume that a prime odd number \(p\) is a Hurwitz prime. Then, \(p|1+l^2+m^2=(1+li+mj)(1-li-mj)\), so \(p\) divides either \(1+li+mj\) or \(1-li-mj\). However, this means that \(p\) divides each coefficient, so it divides 1 too, which is false. Thus, no odd prime is a Hurwitz prime.
Since \(p\) isn't a Hurwitz prime, we can write it as the product of two different Hurwitz quaternions, so \(p=(a+bi+cj+dk)(e+fi+gj+hk)\). By taking modulus, we get that \((a^2+b^2+c^2+d^2)*(e^2+f^2+g^2+h^2)=p^2\), so both of them are equal to \(p\), because, otherwise, their product wouldn't be an integer. If at least one of the sums is composed only of integers, then everything is great. Otherwise, we shall denote \(x=\frac{+/- 1 +/- i +/-j +/- k}{2}\), such that \(p+x\) has only even coefficients. Let \(y=p+x\). Then, \(p=(y^*x-1)(yx^*-1)\), where \(a^*\) is the conjugate of \(a\), while the products belong to the Hurwitz quaternions with integer coefficients, called Lipschitz quaternions. This concludes the proof.
After this, I thought that maybe I could extend the thought processes of this proof to prove easier results. I wanted to prove that any prime number that is a multiple of 4 plus 1 can be written as the sum of two squares. We know that, for such a prime number \(p\), \(-1\) is a quadratic residue, so there is an \(m\) such that \(p|1+m^2\).
I want to prove that \(p\) isn't a Gaussian prime. If \(p\) were a Gaussian prime, then \(p|1+m^2=(i+m)(m-i)\) implies that \(p\) divides both \(m\) and 1, which is a contradiction. Thus, \(p\) isn't a Gaussian prime, so \(p=(a+ib)(c+id)\), with \(a,b,c,d\) integers. By taking the modulus again, we get that both sums are equal to \(p\), which concludes the proof. As a fun fact, any number of the form \(4k+1\) can be written as a product of conjugate Gaussian primes.
Just reading this article makes me wanna study some Number Theory. In Romania, Number Theory isn't treated too much in schools, and, except for the students that participate at TSTs, everyone else stops learning this subject in the 6th grade. Somehow, though, this makes a bit of sense: Number Theory and Algebra go hand in hand in a majestic way, as I just proved, and any interesting proof in this subject would require some Ring Theory. Anyway, I'm gonna go now. Tune in next time for some Ring Theory, Applied Maths, or something completely random!
Cris.
