Matrix problem regarding rational points and a number theory lemma

Eyy there. I was gonna write an article about Markov matrices (which is almost complete, btw), but then, while scrolling through undergrad contests, I found a really neat solution for a problem at the CIIM from last year. The CIIM is a competition for Latin American undergrad students, similar to IMC, Putnam, Seemous and Traian Lalescu. From what I've seen, the problems are a bit harder than those at IMC, but, then again, my opinion isn't necessarily a very competent one. These are the problems (great for you if you know Spanish).



Today, we shall solve the first one (because you know I like matrices). For those that don't know Spanish, here's what it says:

Problem: Prove that there exists a $2*2$ matrix of order 6 with rational entries, such that the sum of its entries is 2018.

Solution: Let $A$ be a matrix with the required properties. Since $A^6=I_2$, we get that $(A-I_2)(A+I_2)(A^2+A+I_2)(A^2-A+I_2)=O_2$. The minimal polynomial of $A$ is, thus, one of those appearing in the decomposition, since $A$ is rational. If $A+I_2=O_2$, then the order of $A$ is 2. If $A-I_2=O_2$, then the order of $A$ is 1. If $A^2+A+I_2=O_2$, then $A^3=I_2$. Thus, the only possibility is that $A^2-A+I_2=O_2$, so $Tr(A)=1$, $det(A)=1$. Let $a, b, c, d$ be the elements of $A$, in that order. Thus, we know that $a+d=1$, and $ad-bc=1$. By writing $d=1-a$, we get that $a(1-a)-bc=1$, so $a-a^2-bc=1$. We want to prove that there exist matrices with these properties such that $a+b+c+d=2018$, which is equivalent to $c=2017-b$. Thus, our end goal is proving that $a-a^2-2017b+b^2=1$ has rational solutions.

Now, we shall write this as the equation of a hyperbola. $\frac{1}{4}+a-a^2-2017b+b^2+2017^2=\frac{5}{4}+2017^2$, and, by multiplying by 4, we get that $(2b-4034)^2-(2a-1)^2=5+4*2017^2$. However, we know that any odd number can be written as a difference of squares ($2k+1=(k+1)^2-k^2$, so there exist two real squares for which we get that conclusion. In fact, we can even calculate $a$ and $b$, but that is left as an exercise for the reader. Since we found that $a$ and $b$ exist, and we found all the rational entries of the matrix according to the rational numbers $a$ and $b$, we conclude that there is indeed a matrix with the required properties.

As a fun fact, initially, I made a calculation mistake. Instead of writing the equation as a hyperbola, I wrote it as the equation of a circle. Thus, I got that $(2b-4034)^2+(2a-1)^2=5+4*2017^2$. I do not know whether the above equation has rational solutions, but, while trying to solve it, I remembered of an old problem from Gazeta Matematica 11/2017, proposed by Marian Andronache, which shall be treated as a lemma:

Lemma: Let $n$ be a natural number bigger than 1 with the sum of its divisors not divisible by 4. Prove that $n$ can be written as the sum of two squares.

Proof: We shall use the well-known 'sum of two squares theorem', which states that a number can be written as a sum of two squares if and only if any prime divisor of the form $4k+3$ appears in the decomposition at an even power. I won't get into details regarding the solution to this, but you can see a proof of this theorem here: https://www.math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Bhaskar.pdf

Getting back to the lemma, let us presume by absurd that $n$ can't be written as a sum of two squares. Thus, there is an odd number of apparitions of a prime number$p_i$ of the form $4m+3$. We know that the sum the divisors is $S(n)=\prod_{k=1}^{w(n)} \frac{p_k^{a_k+1}-1}{p_k-1}$, where $a_k+1$ is the power at which a prime number appears in the decomposition. The term from the product associated with $p_i$ is equal to $1+p_i+p_i^2+...+p_i^{a_i}$, and, because $a_i$ is odd, we get that the sum is a multiple of 4, which contradicts our choice of $n$. The sum is a multiple of 4, because $p^{2k}$ is a multiple of 4 plus 1, while $p^{2k+1}$ is a multiple of 4 plus 3, and we can group the terms together.

I was a bit sad to find out that the above lemma doesn't work on the problem. Indeed, the sum of all divisors of $5+4*2017^2$ is a multiple of 4. It does, however, work on, for example, $1+4*2017^2$. I was just 4 units away. The lemma that I wrote has some interesting applications, and I might search for some problems to solve using it for the next articles. Thinking about that, I haven't posted any number theory problem so far. I might post something after I'm done with the Markov matrices article. Tune in next time for some properties of stochastic matrices, or something completely random!

Cris.