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Q. Which origin-centered circles $C(r)$ (or spheres in dimension $d$) of radius $r < 1$ avoid all rational points of height $\le h$?

A rational point is a point all of whose coordinates are rational numbers. A rational number $x= a/b$ in lowest terms (i.e., gcd$(a,b)=1$) has height $\max \lbrace |a|,|b| \rbrace$. A rational point of height $h$ is a rational point all of whose coordinates are of height $\le h$.

For example, let $h=10$, and $r=\frac{7}{11}$ [Thanks for corrections by NAME_IN_CAPS and Noam Elkies]. Then $C(\frac{7}{11})$, unless I am mistaken, avoids all rational points of height $\le 10$, despite some near misses:


          H10r7/11
Again fix $h=10$, but for $r=\frac{1}{\sqrt{2}}$, $C(r)$ passes through $$\lbrace{ (\frac{1}{10},\frac{7}{10}), (\frac{1}{2},\frac{1}{2}), (\frac{7}{10},\frac{1}{10}) }\rbrace \;:$$
          H10Rsqrt2/2
Perhaps I should phrase the question in the obverse sense:

Q'. Which origin-centered circles $C(r)$ (or spheres in dimension $d$) of radius $r < 1$ include at least one rational point of height $\le h$?

So I am seeking $r$ as a function of $h$. I am unfamiliar with this type of reasoning, but I seek circles that avoid low-height rational points. Thanks!

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    $\begingroup$ Most circles go through no rational point at all. $\endgroup$ – Gerry Myerson Jul 31 '14 at 23:09
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    $\begingroup$ If the denominator of $r^2$ is at least $h^4$ (or $h^{2d}$ in $d$ dimensions), then you're all set. $\endgroup$ – S. Carnahan Jul 31 '14 at 23:13
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    $\begingroup$ Isn't $(3/10)^2+(4/10)^2=(1/2)^2$? $\endgroup$ – NAME_IN_CAPS Jul 31 '14 at 23:26
  • $\begingroup$ @NAME_IN_CAPS: Thanks so much! Indeed I missed that solution. Corrected now. $\endgroup$ – Joseph O'Rourke Aug 1 '14 at 0:41
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    $\begingroup$ Why ignore points such as $\left(0,\pm \frac23\right)$ on $C(2/3)$? $\endgroup$ – Noam D. Elkies Aug 1 '14 at 3:40
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I'm going to interpret the question as:

What is the positive rational number of least height that cannot be written as the sum of $d$ squares of rational numbers of height $<h$?

An easier question might be:

What is the least height of a positive rational number that cannot be written as sum of $d$ squares of rational numbers of height $<h$?

There is a problem for $d=2$, which is some rational numbers can't be written as a sum of rational squares. I believe the smallest height ones are $3$ and $3/5$, so the circle of radius $\sqrt{3}$ and the sphere of radius $\sqrt{3/5}$ might be good answers to this version of the question.

Thus from now on I'm going to assume we do have at least one rational point. For a lower bound, we may use the fact that $a/b$ is a sum of $d$ rational squares if and only if $ab$ is a sum of $d$ integer squares. This follows from the $4$-squares theorem if $d\geq 4$. If $d=2$ or $3$ we can look explicitly at the conditions for $ab$ to be a sum of squares, and note that they are clearly necessary conditions for $a/b$ to be a sum of squares. This lets us write $a/b$ as a sum of squares whose denominator is at most $b$, hence whose numerator is at most $\sqrt{ab}$. This gives a lower bound of $h$.

To get an upper bound close to $h$, let $r^2 =1/p$ for $p$ a prime number slightly larger than $h$ (and, if $d=2$, congruent to $1$ mod $4$, and if $d=3$, not congruent to $7$ mod $8$). Then any rational points which sum to equal $1/p$ must have $p$ somewhere in the denominator of one of them, so must have height at least $p$, hence height $>h$.

So $r^2=1/p$ works quite well. You might find $r^2=a/p$, for $a$ slightly less than $p$ and $a$ a sum of two squares, more aesthetically pleasing.

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  • $\begingroup$ Thank you, Will, for your cogent re-interpretation of my question, and your clear bounds derivations. And especially for the "aesthetically pleasing" final remark! $\endgroup$ – Joseph O'Rourke Aug 1 '14 at 18:35
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This answer does not consider heights of rational numbers, but I hope it still qualifies as related and interesting.

Let $r>0$. The following are equivalent:

  • $C(r)$ avoids all rational points in all dimensions.
  • $r^2$ is irrational.

If $r^2$ is irrational, then $C(r)$ avoids all rational points. If $C(r)$ contained a point $x\in\mathbb Q^d$, then $r^2=x_1^2+\dots+x_d^2\in\mathbb Q$.

Suppose then $r^2=a/b$. Consider the vector $x=(1/b,1/b,\dots,1/b)$ in dimension $d=ab$. Then $x_1^2+\dots+x_d^2=ab\times b^{-2}=r^2$.

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A possible reformulation? If the equation of the circle is rational, then one rational point implies infinitely many. One sees that via lines of rational slope through the point, or the existence of a parametrisation by rational functions (same argument really). Therefore the issue becomes the smallest height of a point, as a function of the radius.

If the equation is not rational, then we are talking about something different. Is that the case you need?

Edit: Well, that wasn't well put. It being a circle, if there is a rational point on it, there is a rational number on the RHS of the standard equation.

I was actually distracted by an old question I was once asked, about printing out all Pythagorean triads in some definite order.

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  • $\begingroup$ That's a nice separation of the problem into two cases: rational equation vs. not. $\endgroup$ – Joseph O'Rourke Aug 1 '14 at 12:00

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