Review question

# How many grid-points can be inside this circle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7626

## Solution

Suppose that $a$, $b$, $c$ are integers such that $a \sqrt{2} + b = c \sqrt{3}.$

1. By squaring both sides of the equation, show that $a = b = c = 0$.

[You may assume that $\sqrt{2}, \sqrt{3}$ and $\sqrt{2/3}$ are all irrational numbers. An irrational number is one which cannot be written in the form $p/q$ where $p$ and $q$ are integers.]

Squaring both sides of the equation gives $2a^2 + 2ab\sqrt{2} + b^2 = 3c^2.$

If $ab \neq 0$ then $\sqrt{2} = \frac{3c^2 - 2a^2 - b^2}{2ab},$ but this is a contradiction because $\sqrt{2}$ is irrational and so cannot be written in the form $\dfrac{p}{q}$ for integers $p$ and $q$.

So we must have either $a=0$ or $b=0$.

If $a=0$, then the initial equation becomes $b=c\sqrt{3}$. Either we have $b=c=0$ or $\sqrt{3}=b/c$. The latter is impossible because $\sqrt{3}$ is irrational.

Similarly, if $b=0$, then the initial equation becomes $a \sqrt{2}=c \sqrt{3}$. Again, we either have $a=c=0$ or $\sqrt{2/3}=c/a$. The latter is impossible as $\sqrt{2/3}$ is irrational.

So either way, we must have $a = b = c = 0$.

1. Suppose now that $m$, $n$, $M$, $N$ are integers such that the distance from the point $(m,n)$ to $(\sqrt{2},\sqrt{3})$ equals the distance from $(M,N)$ to $(\sqrt{2},\sqrt{3})$.

Show that $m=M$ and $n=N$.

Given that the distance from the point $(m,n)$ to $(\sqrt{2},\sqrt{3})$ equals the distance from $(M,N)$ to $(\sqrt{2},\sqrt{3})$, we have $(m-\sqrt{2})^2 + (n-\sqrt{3})^2 = (M-\sqrt{2})^2 + (N-\sqrt{3})^2,$ which after rearranging becomes $2\sqrt{2}(M-m) + (m^2 + n^2 - M^2 - N^2) = 2\sqrt{3}(n-N).$

Now this is in the same form as the given initial equation, with \begin{align*} a &= 2(M-m), \\ b &= (m^2 + n^2 - M^2 - N^2), \\ c &= 2(n-N). \end{align*}

From the first part of the question, we know that $a=b=c=0$, and so we have $2(M-m) = 0 = 2(n-N),$ and hence $m=M$ and $n=N$.

Given real numbers $a$, $b$ and a positive number $r$, let $N(a,b,r)$ be the number of integer pairs $x$, $y$ such that the distance between the points $(x,y)$ and $(a,b)$ is less than or equal to $r$. For example, we see that $N(1.2,0,1.5)=7$ in the diagram below.

1. Explain why $N(0.5,0.5,r)$ is a multiple of $4$ for any value of $r$.

If a particular point $(x,y)$ is within a distance $r$ of $(\frac{1}{2},\frac{1}{2})$, then so are its reflections in the lines $x=\frac{1}{2}$, $y=\frac{1}{2}$, and in both lines.

These are the points $(1-x,y)$, $(x,1-y)$ and $(1-x,1-y)$ respectively.

No points with integer coordinates can lie on these reflection lines.

So the points with integer coordinates that are a maximum distance of $r$ away from $(\frac{1}{2},\frac{1}{2})$ can be split into $4$ sets of equal size by the lines $x=\frac{1}{2}$ and $y=\frac{1}{2}$, as shown in the diagram below.

Hence $N(0.5,0.5,r)$ must be a multiple of $4$.

1. Let $k$ be any positive integer. Explain why there is a positive number $r$ such that $N(\sqrt{2},\sqrt{3},r)=k$.

As $r$ increases, the circle grows and contains more lattice points (points with integer coordinates), without limit.

But in part (ii), we showed that there are no two lattice points equidistant from $(\sqrt{2},\sqrt{3})$, so the circle must only ‘consume’ one lattice point at a time.

Hence $N(\sqrt{2},\sqrt{3},r)$ can take any positive integer value.