2
$\begingroup$

Suppose we are given integers $k,c$ such that $k=1+c^2$.

Let $n$ be an odd integer and suppose that $k^n=a_i^2+b_i^2$ for distinct positive integers $a_i<b_i$ and $i\le d$. That is, there are $d$ different ways to express $k^n$ as a sum of two squares.

For instance, $(a_1,b_1)=(k^{(n-1)/2},ck^{(n-1)/2})$ is a valid pair.

What can be said about $\max b_i$? Are there good bounds (as a function of $c,n$) on the magnitude of largest possible square when writing an integer as a sum of two squares?


Addendum: As mentioned below, we can rephrase this problem in terms of the irrationality measure of $\arctan(1/c)/\pi$. I'm having a lot of trouble finding results on irrationality measures of values of inverse trigonometric functions in general, but I could be missing something.

$\endgroup$
1
  • $\begingroup$ if $\theta=\arctan 1/c$, we need an odd multiple of $\theta$ which is close to 0 modulo $2\pi$, this may be tricky $\endgroup$ May 17, 2022 at 15:13

1 Answer 1

1
$\begingroup$

Rather than discuss $\max b_{i}$, I'll discuss the equivalent question of bounding $\min a_{i}$. The ABC conjecture implies that for all $\epsilon > 0$, $\min a_{i} \gg (c^{2}+1)^{n/2 - 1 - \epsilon}$. This is because if we have a solution to $a^{2} + b^{2} = (c^{2}+1)^{n}$ with $a \ll (c^{2} + 1)^{n/2 - 1 - \epsilon}$, set $A = a^{2}$, $B = b^{2}$ and $C = (c^{2}+1)^{n}$. Assume for simplicity that $\gcd(a,b) = 1$. (It doesn't change much if $\gcd(a,b) > 1$.) Then $C \ll {\rm rad}(ABC)^{1+\delta}$ for all $\delta > 0$. This gives $$ (c^{2}+1)^{n} < (ab(c^{2}+1))^{1+\delta} \ll ((c^{2}+1)^{n/2 - 1 - \epsilon} (c^{2}+1)^{n/2} (c^{2}+1))^{1+\delta} = ((c^{2}+1)^{n-\epsilon})^{1+\delta} $$ which is a contradiction if $\delta < \frac{\epsilon}{n-\epsilon}$.

For $n = 3$, it is possible to construct a sequence of integers $c$ which getse close to this bound. In particular, let $$ c_{k} = \frac{(2 + \sqrt{3})^{k} - (2 - \sqrt{3})^{k}}{\sqrt{3}} \quad a_{k} = \frac{(2 + \sqrt{3})^{k} + (2 - \sqrt{3})^{k}}{2}. $$ It is easy to see that $a_{k}, c_{k} \in \mathbb{Z}$, $c_{k}$ is even, and a somewhat tedious calculation shows that $$ (c_{k}^{2} + 1)^{3} = a_{k}^{2} + \left(c_{k}^{3} + \frac{3}{2} c_{k}\right)^{2}. $$ In particular $\min a_{i} \leq a_{k} \approx \frac{\sqrt{3}}{2} c_{k}$.

One could hope to generalize this construction by finding $c_{k}^{2} + 1 = d_{k}^{2} + e_{k}^{2}$, and $\frac{e_{k}}{c_{k}} \approx \sin\left(\frac{\pi}{2n}\right)$. Letting $\frac{e_{k}}{\sqrt{d_{k}^{2} + e_{k}^{2}}} = \sin(\theta_{k})$, this makes $$ (c_{k} + i)^{n} = (d_{k} + ie_{k})^{n} \approx (c_{k}^{2} + 1)^{n/2} \left(\cos(\theta_{k}) + i \sin(\theta_{k})\right)^{n} = (c^{2} + 1)^{n/2} \left(\cos(n \theta_{k}) + i \sin(n \theta_{k}\right)) $$ and $\sin(n \theta_{k}) \approx \sin(\pi/2) = 1$. This boils down to finding points on the hyperboloid $x^{2}+y^{2} = z^{2} + 1$ that lie close to the line $y = \sin\left(\frac{\pi}{2n}\right) z$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.