Saturday 18 November 2023

Simon Marais 2023B4 revisited

(SM2023, B4) Let $n$ be a non-square number. 

(a) Find all pairs of natural numbers $(a,b)$ so that $r^a+\sqrt{n}, r^b+\sqrt{n}$ are both rational for some positive real $r$.
(b) Find all pairs of natural numbers $(a,b)$ so that $r^a+\sqrt{n}, r^b+\sqrt{n}$ are both rational for some real $r$. This is an open question currently.

When I first drafted my quick comment I simply omitted this question because a Q4 is a Q4 and has to be respected, so I only left one or two sentences there. Later on I found this one of interest so I dived in and wrote a simple answer. But then some gap in the solution had been lingering in my mind, and unsurprisingly I missed out something important so the answer is not fully correct. But I feel like I am close to the answer since the math behind is nothing overly complicated.

Below is my later attempt. There is a gap still, not recommended for readers.


Obviously, every $a=b$ pair is a solution because you can easily make up $r = (n-\sqrt{n})^{1/a}$. Next, if $(a,b)$ is a solution then so as its multiples $(ka,kb)$ because you can take $r' = r^{1/k}$. 

As a result we can now assume that $a,b$ are coprime. As $r^a, r^b \in \mathbb{Q}[\sqrt{n}]$, we know that $r\in \mathbb{Q}[\sqrt{n}]$ as well. I think this is well known -- but in case that is not, think about Euclidean algorithm (recall that a field is Euclidean...or even simpler the algorithm where you find gcd) that you can divide each other until you reach $r^1 = r$. 

Now write $r= p-q\sqrt{n}$ for some positive rational $p,q$ (the proofs are similar for different sign combination. Like you can write $r = p+q\sqrt{n}$ can the argument below proceeds the same). If you still remember your linear algebra lesson you can solve the recurrence by writing $p_k-q_k \sqrt{n} = (p-q\sqrt{n})^k$ for rational $(p_k), (q_k)$. For $v_k = (p_k,q_k)^t$, we have $v_k = A^{k}v_0$ where $A = \begin{bmatrix}p & nq\\ q & p\end{bmatrix}$.

The general formula for $p_k,q_k$ can be easily obtained by either method of eigenvectors or if you know how expand binomial powers. Anyway we have: 
$p_k = \frac{1}{2}((p+q\sqrt{n})^k+(p-q\sqrt{n})^k)$
$q_k = \frac{1}{2\sqrt{n}}((p+q\sqrt{n})^k-(p-q\sqrt{n})^k)$.
Notice that the formula is the same for $(p+q\sqrt{n})^k$. For negative $p,q$ the sign just alternates. 

Clearly $q_k$ is an alternating almost exponential series. More precisely we expect $q_{2k}-1 \approx q(p+q\sqrt{n})^{2k-2}$ and $q_{2k}\approx 2pq(p+q\sqrt{n})^{2k-2}$. The two subsequences are surely strictly monotonic and would not give us any non-trivial answer. The difficult task is however to find $k,k'$ so that $q_{2k} = q_{2k'-1}$.

A few approaches are there: first you may want to show that $q_n$ may not even be an integer if $p,q$ are (non-integral) fractions past a certain $k$. Secondly you may consider something even stronger that $q_k$ can't be 1 for largr $k$ because exponential sequences are either diverging or goes to zero, but the problem is you can always make very marginal case where $(p+q\sqrt{n})$ is very close to 1 (e.g. using continued fractions) so that the 'rate of divergence' is not properly bounded.

The problem for us is (1) we can't calculate terms manually for anything past $k=4$ -- even that is nasty enough as you can see below, and (2) what is the meaning of (b) being unsolved? Most arguments, if we assume that $r\in \mathbb{Q}[\sqrt{n}]$, does not really care about its positivity. In fact, we care if $(p-q\sqrt{n})$ being negative or not more so $(1-2\sqrt{2})^k$ is easier to handle than $(1+2\sqrt{2})^k$!

My approach is basically the observation that the last $k$ where $q_k$ can easily be made as an integer when $p,q\notin \mathbb{Z}$ is $q_4$ with where $(\frac{1}{2} + \frac{3}{2}\sqrt{3}) = \frac{223}{4} + 21\sqrt{3}$. One suspects that for larger $k$ the expansion looks like $q_k = kp^{k-1}q + cnp^{k-3}q^3 + c'n^2q^5(...)$, where the $n^2$ is causing problems when we want to solve integrality or doing mod checks. But at the end the approach does not distinguish between (a) and (b). Now I am really confused...and I see nothing wrong in the Euclidean argument too. 

Still, let us try to check the lower $q_k$'s:

$q_1 = q$
$q_2 = 2pq$
$q_3 = 3p^2q + nq^3$
$q_4 = 4pq(p^2+nq^2)$

As claimed, $q_1\neq q_3$ and $q_2\neq q_4$ so we only need to solve the 12,14,23,34 pairs.

$q_1=q_2=1$: this is easy with $p = \frac{1}{2}$ and $q=1$. Possibly a solution that students would have noticed without going through all these hassle.

$q_2=q_3=1$: Substitution gives the quintic equation $8p^4 - 4p^3 + n = 0$ with discriminant $256n^2(512n-27)$. By checking $(512n-27)$ mod 8 we know this is a non-square so that gives no rational solution.

$q_1=q_4=1$: Substitution gives the equation $4p(p^2+n)-1 = 4p^3 + 4np-1 = 0$. Surely one real root but the discriminant of a cubic equation doesn't tell much about rationality. Instead we convert $p$ into a fraction $p = \frac{s}{t}$ so $4p(p^2+n)$ becomes $\frac{4s(s^2+nt^2)}{t^3}$ for some $(s,t)=1$ but then we require $t^2\mid s^2$ which is absurd.

$q_3=q_4=1$: this is of course the hardest...but the technique never changes. From $q_3=1$ we obtain $q = (4p-1)/8p^3$. Substitution into $q_4=1$ gives a horrible degree 9 equation 
$3p^2((4p-1)/(8p^3)) + n((4p-1)/(8p^3))^3$
$ = (n(4p-1)^3 - 3\times 64p^2 + 3\times 256p^3)/512p^9 = 1$. 
Looks horrible but all we need is to take mod $p^2$ which eliminates the last two terms. Since $(p,4p-1)=1$ that forces $n$ to be non-square-free, hence the contradiction. So $(a,b) = (k,k)$ or $(k,2k)$ are all the possible pairs.

Other than discriminant checking, the main trick is to force a term to be multiple of $p^2$ (or anything else) if you want the sum to be a multiple of $n^2$ and so does the rest of the terms, then we can apply the square-free assumption. I believe this can be applied for higher $q_k$'s because we can check higher powers of $p$ against the $n^2$ factor. But clearly (b) being open does not support that.

So, basically the sure answers are in forms of $(a,b) = (k,k)$ or $(k,2k)$. But are there more? I don't know.

The more I think about this question the more I like it with so many different scattered technique that were used. This is the type of question we would like to see more. I would not hide the fact that my first attempt is a miserable fail as I overlooked a large part of it. I wrote that we look at $A^2$ again we know from there that the denominator would sure to stack up every 2 steps, but what about consecutive terms $q_k$ and $q_{k+1}$? I stared at what I wrote before I finally realized what should be corrected...but nope I am probably not getting 7 this time.

Just food for thought...what if we are now in $\mathbb{Q}[\sqrt{-n}]$, or that $(a,b)$ are simply integers instead? 

No comments:

Post a Comment