1

The exercise is the following:

Evaluate the limit $\lim \limits_{R \to \infty} \frac{n(1)+...n(R)}{R^{3/2}}$, given that $R>0$ is an integer and $n(R)$ is the number of integer solutions $(x,y,z) \in \mathbb{Z}^3$ to $2x^2+3y^2+5z^2=R$.

This is an olympiad problem, so I think it might rely on some trick or clever insight. I took a look at some small cases and it looks like $\{\frac{1}{2}\sqrt2,\sqrt2,\sqrt2,\frac{3}{2} \sqrt2, ... \}$. I also though about an ellipsoid of the form: $\frac{x^2}{(\sqrt{R/2})^2}+\frac{y^2}{(\sqrt{R/3})^2}+\frac{z^2}{(\sqrt{R/5})^2} = 1$, then writing $(x,y,z)$ as $(\sqrt{R/2}\sin{u}\cos{v},\sqrt{R/3}\sin{u}\sin{v},\sqrt{R/5}\cos{u})$. I am really lost at this problem, so any insight on it and also on what should I read to prepare myself to similar problems would be of great value.

Alexandre Tourinho
  • 829
  • Unless a typo, it says $(x,y,z) \in \mathbb{Z}^3$ and that would restrict number of solutions – Math Lover Jan 05 '22 at 17:47
  • Not a typo, that is indeed a restriction to the integers, so it seems to involve some number theory, but I am not sharp enough to see what follows – Alexandre Tourinho Jan 05 '22 at 17:50
  • 2
    You need to count/estimate the number of integer points inside the ellipsoid. – richrow Jan 05 '22 at 17:58
  • Except $R = 1, 4, 6$ and may be a few others, I think $2x^2+3y^2+5z^2$ generates all integers $R$. So what you need to check if there is unique solution for every $R$... that may be the point – Math Lover Jan 05 '22 at 18:02
  • 1
    also see https://math.stackexchange.com/questions/2149242/how-to-show-that-2x-3y-5z-generate-all-integers-greater-than-2 – Math Lover Jan 05 '22 at 18:04
  • @MathLover In many cases the solution is not unique even excluding negative values, starting with $5 = 2(\pm 1)^2 + 3(\pm 1)^2 + 5(0)^2 = 2(0)^2 + 3(0)^2 + 5(\pm 1)^2$ – aschepler Jan 05 '22 at 18:08
  • @aschepler yes you are right. My idea was to see if it somehow restricted the number of solutions but I may be on a wrong track. I should not have said unique in any case. – Math Lover Jan 05 '22 at 18:10
  • @richrow's hint about the number of points inside the ellipsoid gives a quick way to find what the limit must be if it does exist. Proving that the sequence does in fact converge to that value is a little bit trickier, but might not be needed in the original contest context, depending on how the answer should be specified/used. – aschepler Jan 05 '22 at 18:20
  • Yeah @aschelper this richrow idea of estimating points inside the ellipsoid looks nice, but not so quick for me. Can you elaborate? I can't see how to deal with the summation on the numerator of the limit – Alexandre Tourinho Jan 05 '22 at 18:31
  • 2
    Consider the set $S(R)={(x,y,z)\in\mathbb{Z}^3\colon 2x^2+3y^2+5z^2\le R}$. Prove that the cardinality of $S(R)$ equals $1+n(1)+\ldots+n(R)$. – richrow Jan 05 '22 at 19:00
  • I agree that for large R the sum is approximately the number of points in the positive octant of the ellipsoid, which, if I am not mistaken, is equal to $\frac{\pi}{6\sqrt{30}} R^{\frac{3}{2}}$ – blamocur Feb 02 '22 at 23:07

1 Answers1

2

First, note $n(1) + \cdots + n(R)$ is the number of integer triples $(x,y,z)$ such that $2x^2+3y^2+5z^2 \leq R$. Let's call this count $N(R) = n(1) + \cdots + n(R)$.

We can also consider the inequality $2x^2+3y^2+5z^2 \leq R$ as a statement about four real variables, not restricted to integers only, but requiring $R \geq 0$. In this case, it describes the surface and interior of an ellipsoid. We can find its semi-axis lengths by converting to the form:

$$ \left(\frac{x}{\sqrt{R/2}}\right)^2 + \left(\frac{y}{\sqrt{R/3}}\right)^2 + \left(\frac{z}{\sqrt{R/5}}\right)^2 \leq 1 $$

Name the semi-axis lengths $a=\sqrt{R/2}, b=\sqrt{R/3}, c=\sqrt{R/5}$, and the bounding equality has standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$.

Intuitively, we expect the number of integer points $N(R)$ satisfying the inequality to be approximately the volume of the ellipsoid, $V(R) = \frac{4}{3}\pi a b c = \frac{2}{45} \sqrt{30} \pi R^{3/2}$. This would make the limit of $\frac{N(R)}{R^{3/2}}$ equal to $\frac{2}{45} \sqrt{30} \pi$.

Now to rigorously show $N(R)$ does converge to that limit. With each integer triple $(x,y,z)$ with $x,y,z \in \mathbb{Z}$, associate the unit cube $c(x,y,z) \subset \mathbb{R}^3$:

$$ c(x,y,z) = \left[x-\frac 12, x+\frac 12\right] \times \left[y-\frac 12, y+\frac 12\right] \times \left[z-\frac 12, z+\frac 12\right] $$

Or equivalently, $$ c(x,y,z) = \left\{(x', y', z') \in \mathbb{R}^3 : |x'-x| \leq \frac 12 \mbox{ and } |y'-y| \leq \frac 12 \mbox{ and } |z'-z| \leq \frac 12 \right\} $$

These each have volume $1$ and their overlaps have volume $0$, so $N(R)$ is the volume of the set of all cubes associated with the integer solutions to $2x^2 + 3y^2 + 5z^2 \leq R$. If $(x,y,z)$ is one of these solutions, then

$$ 2x^2+3y^2+5z^2 \leq R $$ $$ |x| \leq \sqrt{R/2} \quad |y| \leq \sqrt{R/3} \quad |z| \leq \sqrt{R/5} $$

Let $(x',y',z') \in c(x,y,z)$. Define $F(x',y',z')$: $$ \begin{align*} F(x',y',z') &= 2x'^2 + 3y'^2 + 5z'^2 \\ F(x',y',z') &\leq 2 \left(|x|+\frac{1}{2}\right)^2 + 3 \left(|y|+\frac{1}{2}\right)^2 + 5 \left(|z|+\frac{1}{2}\right)^2 \\ F(x',y',z') &\leq 2x^2 + 3y^2 + 5z^2 + 2|x| + 3|y| + 5|z| + \frac{2+3+5}{4} \\ F(x',y',z') &\leq R + \sqrt{2R} + \sqrt{3R} + \sqrt{5R} + \frac{5}{2} \\ F(x',y',z') &< R + (2+2+3)\sqrt{R} + \frac{5}{2} \\ F(x',y',z') &< R + 7\sqrt{R} + 3 \\ \end{align*} $$

So letting $R_U = R+7\sqrt{R}+3$, $c(x,y,z)$ is within the interior of a larger ellipsoid solid:

$$ 2x'^2+3y'^2+5z'^2 < R_U $$

Since all the cubes $c(x,y,z)$ with $2x^2+3y^2+5z^2 \leq R$ are inside this ellipsoid, $N(R)$ is less than its volume:

$$ N(R) < \frac{2}{45}\sqrt{30} \pi R_U^{3/2} $$ $$ \frac{N(R)}{R^{3/2}} < \frac{2}{45}\sqrt{30} \pi \left(\frac{R+7\sqrt{R}+3}{R}\right)^{3/2} = \frac{2}{45}\sqrt{30} \pi \left(1+\frac{7}{\sqrt{R}}+\frac{3}{R}\right)^{3/2} $$

Now suppose $x,y,z$ are integers such that $2x^2+3y^2+5z^2 > R$, and $(x',y',z') \in c(x,y,z)$. If $x \neq 0$ then

$$ |x'| \geq |x|-\frac{1}{2} > 0 $$ $$ x'^2 \geq x^2-|x|+\frac{1}{4} \geq x^2-|x| $$

Or if $x=0$, it's still true that $x'^2 \geq x^2-|x|$, so this holds for any integer $x$. Similarly, we have $y'^2 \geq y^2-|y|$ and $z'^2 \geq z^2-|z|$.

$F(x',y',z')$ is the same formula as above:

$$ \begin{align*} F(x',y',z') &= 2x'^2+3y'^2+5z'^2 \\ F(x',y',z') &\geq 2x^2+3y^2+5z^2 - 2|x|-3|y|-5|z| \\ F(x',y',z') &> R - 2\sqrt{R/2} - 3\sqrt{R/3} - 5\sqrt{R/5} \\ F(x',y',z') &> R - \sqrt{2R} - \sqrt{3R} - \sqrt{5R} \\ F(x',y',z') &> R - 7\sqrt{R} \end{align*} $$

So letting $R_L = R - 7\sqrt{R}$, all the cubes $c(x,y,z)$ with $2x^2+3y^2+5z^2 > R$ are within the exterior of a smaller ellipsoid solid:

$$ 2x'^2+3y'^2+5z'^2 > R_L $$

So the ellipsoid is entirely inside the union of the $N(R)$ cubes where $2x^2+3y^2+5z^2 \leq R$, and $N(R)$ is larger than its volume:

$$ N(R) > \frac{2}{45}\sqrt{30} \pi R_L^{3/2} $$ $$ \frac{N(R)}{R^{3/2}} > \frac{2}{45}\sqrt{30} \pi \left(\frac{R-7\sqrt{R}}{R}\right)^{3/2} = \frac{2}{45}\sqrt{30} \pi \left(1-\frac{7}{\sqrt{R}}\right)^{3/2} $$

Finally we can apply the squeeze theorem. For any non-negative integer $R$,

$$ \frac{2}{45}\sqrt{30} \pi \left(1-\frac{7}{\sqrt{R}}\right)^{3/2} < \frac{N(R)}{R^{3/2}} < \frac{2}{45}\sqrt{30} \pi \left(1+\frac{7}{\sqrt{R}}+\frac{3}{R}\right)^{3/2} $$

Both bounds approach $\frac{2}{45}\sqrt{30} \pi$ as $R \to \infty$, so the limit does indeed exist with that value.

aschepler
  • 11,035