Let $\alpha,\beta$ be positive real numbers and $N_{\alpha,\beta}(x)$ denote the number of integer pairs $(m,n)$ satisfying

\[\alpha m+\beta n\le x,\quad m,n\ge0.\tag1\]

Then it follows from some geometric motivations that $N_{\alpha,\beta}(x)$ should be well approximated by ${x^2\over2\alpha\beta}$, the area of the triangle. In this article, we prove that when

\[P(x)={x^2\over2\alpha\beta}+{x\over2\alpha}+{x\over2\beta},\]

we have

\[N_{\alpha,\beta}(x)-P(x)=o(x)\tag2\]

for irrational $\alpha/\beta$, and when $\alpha/\beta$ is rational, there exists $K>0$ such that for all $X>0$, there exist $x,x'>X$ satisfying

\[N_{\alpha,\beta}(x)-P(x)>Kx\tag3\]

and

\[N_{\alpha,\beta}(x')-P(x')<-Kx'.\tag4\]

The methods used in this article is adapted from G. H. Hardy's book on Ramanujan1.

Expression for the error term

Error terms in lattice counting problems often involve the difference of $x$ and $\lfloor x\rfloor$. Since $x-\lfloor x\rfloor$ is a periodic function taking values from $[0,1]$, it is expected that sums over it should be approximated by the sum over its “average value” $\frac12$, so we define a new function for convenience:

\[\rho(x)=x-\lfloor x\rfloor-\frac12.\tag5\]

As a result, there is

\[\begin{aligned} N_{\alpha,\beta}(x) &=\sum_{0\le m\le x/\alpha}\sum_{0\le n\le(x-\alpha m)/\beta}1=\sum_{0\le m\le x/\alpha}\left(1+\left\lfloor x-\alpha m\over\beta\right\rfloor\right) \\ &=\sum_{m=0}^{\lfloor x/\alpha\rfloor}\left(\frac12+{x-\alpha m\over\beta}\right)+\sum_{m=0}^{\lfloor x/\alpha\rfloor}\rho\left(x-\alpha m\over\beta\right). \end{aligned}\]

Let $x/\alpha-\lfloor x/\alpha\rfloor=w$, so the first term becomes

\[\begin{aligned} \sum_{m=0}^{\lfloor x/\alpha\rfloor}\left(\frac12+{x-\alpha m\over\beta}\right) &=\frac12\left\lfloor\frac x\alpha\right\rfloor+\frac x\beta\left\lfloor\frac x\alpha\right\rfloor-{\alpha\over2\beta}\left\lfloor\frac x\alpha\right\rfloor\left(\left\lfloor\frac x\alpha\right\rfloor+1\right) \\ &={x\over2\alpha}+{x^2\over\alpha\beta}-{xw\over\beta}-{\alpha\over2\beta}\left(\frac x\alpha-w\right)^2+{\alpha\over2\beta}\left\lfloor\frac x\alpha\right\rfloor+O(1) \\ &={x\over2\alpha}+{x^2\over\alpha\beta}-{xw\over\beta}-{\alpha\over2\beta}\left({x^2\over\alpha^2}-{2wx\over\alpha}+w^2\right)+{x\over2\beta}+O(1) \\ &={x^2\over2\alpha\beta}+{x\over2\alpha}+{x\over2\beta}+O(1)=P(x)+O(1). \end{aligned}\]

Combining everything, we see that

\[N_{\alpha,\beta}(x)-P(x)=S_{\alpha,\beta}(x)+O(1),\tag6\]

where

\[S_{\alpha,\beta}(x)=\sum_{m=0}^{\lfloor x/\alpha\rfloor}\rho\left(x-\alpha m\over\beta\right).\tag7\]

Estimates of $S_{\alpha,\beta}(x)$ when $\alpha/\beta$ is irrational.

Notice that $S_{\alpha,\beta}(x)=S_{\alpha\beta^{-1},1}(x\beta^{-1})$, so we assume $\beta=1$ and $\alpha$ irrational for convenience, so (7) simplifies into

\[S_{\alpha,1}(x)=\sum_{m=0}^{\lfloor x\rfloor}\rho(x-\alpha m).\]

Let $\frac pq={p_n\over q_n}$ be an n'th continued fraction convergent to $\alpha$. Then it follows from the properties of continued fractions that $p,q$ are coprime and

\[\alpha=\frac pq+(-1)^n\varepsilon,\quad 0<\varepsilon<{1\over q^2}.\tag8\]

Write $\lfloor x\rfloor=rq+s$ for some $0\le s<q$, so there is

\[\begin{aligned} S_{\alpha,1}(x) &=\sum_{m=0}^{rq-1}\rho(x-\alpha m)+O(s) \\ &=\sum_{\mu=0}^{r-1}\underbrace{\sum_{\nu=0}^{q-1}\rho(x-\alpha\mu q-\alpha\nu)}_{S^{(\mu)}(x)}+O(s). \end{aligned}\tag9\]

Bounds for $S^{(\mu)}(x)$

Write $\theta_\mu=x-\alpha\mu q$ and choose $t\in\mathbb Z$ such that

\[\theta_\mu=\frac tq+(-1)^n\delta,\quad 0\le\delta<\frac1q.\tag{10}\]

Then it follows from (8) and $0\le\nu<q$ that

\[\theta_\mu-\alpha\nu-{t-p\nu\over q}=(-1)^n(\delta+\nu\varepsilon)\in\left(-\frac1q,\frac1q\right).\tag{11}\]

Because $\gcd(p,q)=1$, $\nu\mapsto p\nu$ permutes $\mathbb Z/q\mathbb Z$. As a result,

\[r_\nu={t-p\nu\over q}-\left\lfloor t-p\nu\over q\right\rfloor\]

takes each of $0,1/q,2/q,\dots,(q-1)/q$ exactly once when $\nu$ runs through $0,1,2,\dots,q-1$. It follows from (11) that

\[\lfloor\theta_\mu-\alpha\nu\rfloor=\left\lfloor t-p\nu\over q\right\rfloor\]

when $r_\nu\ne0$ and when $r_\nu=0$,

\[\lfloor\theta_\mu-\alpha\nu\rfloor=\left\lfloor t-p\nu\over q\right\rfloor+g\]

for some $g\in\lbrace 0,-1\rbrace$ according to whether there is an integer between $\theta_\mu-\alpha\nu$ and $t-p\nu\over q$. Since such exceptional $r_\nu$ only occurs once, we have

\[\begin{aligned} S^{(\mu)}(x) &=\sum_{\nu=0}^{q-1}\left(\theta_\mu-\alpha\nu-\frac12\right)-\sum_{\nu=0}^{q-1}\lfloor\theta_\mu-\alpha\nu\rfloor \\ &=\left(\theta_\mu-\frac12\right)q-\frac\alpha2 q(q-1)-\sum_{\nu=0}^{q-1}\left\lfloor t-p\nu\over q\right\rfloor-g \\ &=\left(\theta_\mu-\frac12\right)q-\frac\alpha2 q(q-1)+\sum_{\nu=0}^{q-1}r_\nu-\sum_{\nu=0}^{q-1}{t-p\nu\over q}-g \\ &=\left(\theta_\mu-\frac12\right)q-\frac\alpha2 q(q-1)+\sum_{\nu=0}^{q-1}{\nu\over q}-\sum_{\nu=0}^{q-1}\left(\frac tq-\frac pq\cdot\nu\right)-g \\ &=\left(\theta_\mu-\frac12-\frac tq\right)q-\frac12\left(\alpha-\frac pq\right)q(q-1)+{q-1\over2}-g \\ &=\left(\theta_\mu-\frac tq\right)q-\frac12\left(\alpha-\frac pq\right)q(q-1)-\frac12-g. \end{aligned}\]

Combined with (8) and (10), we have

\[\vert S^{(\mu)}(x)\vert\le 1+{q(q-1)\over 2q^2}+\frac12+\vert g\vert<3.\]

Plugging this into (9), we obtain

\[S_{\alpha,1}(x)=\sum_{\mu=0}^{r-1}O(1)+O(s)=O(r)+O(s).\]

Since $r\le x/q$ and $0\le s<q$, we conclude that there is some fixed $A>0$ satisfying

\[\vert S_{\alpha,1}(x)\vert\le A\left(\frac xq+q\right).\]

When $\alpha$ is irrational, it follows from the theory of continued fractions that there are infinitely many pairs of coprime integers $(p,q)$ satisfying (8), so for any $\varepsilon>0$ we can choose $q>2A\varepsilon^{-1}$, so when $x>2Aq\varepsilon^{-1}$, there is

\[\vert S_{\alpha,1}(x)\vert<\frac\varepsilon2x+\frac\varepsilon2x=\varepsilon x.\]

This indicates that when $\alpha/\beta$ is irrational and $x\to+\infty$, there is

\[S_{\alpha,\beta}(x)=o(x).\]

Combining this with (6) completes the proof of (2).

Oscillation of $S_{\alpha,\beta}(x)$ when $\alpha/\beta$ is rational

The elementary approach presented below is my original idea, which is considerably simpler than the complex-analytic method in Hardy's book.

When $\alpha/\beta$ is rational, there are only finitely many coprime pairs $(p,q)$ satisfying (8), so the method above can only give us $S_{\alpha,\beta}(x)=O(x)$. In this section, we prove that no further improvements can be made.

Since $S_{\alpha,\beta}(x)=S_{\alpha\lambda,\beta\lambda}(\lambda x)$ for all $\lambda>0$, we assume without loss of generality that $\alpha$ and $\beta$ are coprime integers.

Write $\lfloor x/\alpha\rfloor=r\beta+s$ for some $0\le s<\beta$, so it follows from (7) and the fact that $\rho(x+1)=\rho(x)$,

\[\begin{aligned} S_{\alpha,\beta}(x) &=\sum_{m=0}^{r\beta-1}\rho\left(x-\alpha m\over\beta\right)+O(1) \\ &=\sum_{\mu=0}^{r-1}\sum_{\nu=0}^{\beta-1}\rho\left(\frac x\beta-{\alpha(\mu\beta+\nu)\over\beta}\right)+O(1) \\ &=\sum_{\mu=0}^{r-1}\sum_{\nu=0}^{\beta-1}\rho\left(\frac x\beta-\alpha\mu-{\alpha\nu\over\beta}\right)+O(1) \\ &=r\sum_{\nu=0}^{\beta-1}\rho\left(\frac x\beta-{\alpha\nu\over\beta}\right)+O(1). \end{aligned}\]

Since $\nu\mapsto-\alpha\nu$ permutes $\mathbb Z/\beta\mathbb Z$, we can rewrite the remaining sum into

\[\begin{aligned} \sum_{\nu=0}^{\beta-1}\rho\left(\frac x\beta+{\alpha\nu\over\beta}\right) &=\sum_{\nu=0}^{\beta-1}\rho\left(\frac x\beta+\frac\nu\beta\right) \\ &=\sum_{\nu=0}^{\beta-1}\left(\frac x\beta+\frac\nu\beta-\frac12\right)-\sum_{\nu=0}^{\beta-1}\left\lfloor\frac x\beta+\frac\nu\beta\right\rfloor \\ &=x-\frac\beta2+\frac1\beta\frac\beta2(\beta-1)-\sum_{\nu=0}^{\beta-1}\left\lfloor\frac x\beta+\frac\nu\beta\right\rfloor \\ &=x-\frac12-\sum_{\nu=0}^{\beta-1}\left\lfloor\frac x\beta+\frac\nu\beta\right\rfloor=x-\lfloor x\rfloor-\frac12, \end{aligned}\]

where the last equality invokes Hermite's identity for floor function.

Since $r={x\over\alpha\beta}+O(1)$, we conclude that

\[S_{\alpha,\beta}(x)={x\over\alpha\beta}\left(x-\lfloor x\rfloor-\frac12\right)+O(1).\tag{12}\]

Because $x-\lfloor x\rfloor-\frac12$ oscillates between $-\frac12$ and $\frac12$ as $x\to+\infty$, plugging (12) into (6) effectively finishes the proof of (3) and (4).

Conclusion

In this article, we performed a systematic study of the error $N_{\alpha,\beta}(x)-P(x)$. By invoking the theory of continued fractions, we proved that the error is of order $o(x)$ when $\alpha/\beta$ is irrational. Subsequently, we gave an elementary proof that the error is oscillating with order of magnitude $x$ when $\alpha/\beta$ is rational.

  1. Hardy, G. H. (1940). Ramanujan: Twelve lectures on subjects suggested by his life and work. Cambridge University Press. ↩