Review question

Can we find this circle's integer and non-integer rational points? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9072

Solution

1. The point with coordinates $(a,b)$, where $a$ and $b$ are rational numbers, is called:
• an integer rational point if both $a$ and $b$ are integers;

• a non-integer rational point if neither $a$ nor $b$ is an integer.

1. Write down an integer rational point and a non-integer rational point on the circle $x^2+y^2=1$.

An integer rational point? This is easy - $(0,1)$, for example.

A non-integer rational point? We can use Pythagoras’ Theorem, which tells us that a $3-4-5$ triangle is right-angled.

Thus a $3/5-4/5-1$ triangle is right-angled, and $\left(\frac{3}{5},\frac{4}{5}\right)$ is a non-integer point on the circle.

1. Write down an integer rational point on the circle $x^2+y^2=2$. Simplify $(\cos\theta+\sqrt{m}\sin\theta)^2+(\sin\theta-\sqrt{m}\cos\theta)^2$ and hence obtain a non-integer rational point on the circle $x^2+y^2=2$.

Once again, finding an integer rational point on the circle is easy: $(1,1)$, for example.

Multiplying out, we find that \begin{align*} (\cos\theta &+ \sqrt{m}\sin\theta)^2+(\sin\theta-\sqrt{m}\cos\theta)^2\\ &=\cos^2 \theta + 2\sqrt{m}\cos\theta\sin\theta+m\sin^2\theta+\sin^2\theta - 2\sqrt{m}\cos\theta\sin\theta + m\cos^2\theta\\ &=(1+m)\left(\cos^2\theta+\sin^2\theta\right)=1+m. \end{align*}

So let’s put $m=1$, and the above becomes $\left(\cos\theta+\sin\theta\right)^2+\left(\sin\theta-\cos\theta\right)^2=2$, which looks like the equation of our circle. So the point $(x,y)=(\cos\theta +\sin\theta, \sin\theta-\cos\theta)$ must lie on the given circle.

So if we choose $\theta$ to be the angle in the same $3/5-4/5-1$ right-angled triangle as before, we find \begin{align*} x&=\cos\theta+\sin\theta = \frac{4}{5}+\frac{3}{5}=\frac{7}{5}\\ y&=\sin\theta-\cos\theta=\frac{3}{5}-\frac{4}{5}=-\frac{1}{5}. \end{align*}

Check: $\left(-\dfrac{1}{5}\right)^2+\left(\dfrac{7}{5}\right)^2 = 2$.

Thus a non-integer rational point on the circle $x^2+y^2=2$ is $\left(\frac{7}{5},-\frac{1}{5}\right)$.

1. The point with coordinates $(p+\sqrt{2}q,r+\sqrt{2}s)$, where $p$, $q$, $r$ and $s$ are rational numbers, is called:
• an integer $2$-rational point if all of $p$, $q$, $r$ and $s$ are integers;

• a non-integer $2$-rational point if none of $p$, $q$, $r$ and $s$ is an integer.

1. Write down an integer $2$-rational point, and obtain a non-integer $2$-rational point, on the circle $x^2+y^2=3$.

This time we let $m=2$. Thus $(\cos\theta+\sqrt{2}\sin\theta)^2+(\sin\theta-\sqrt{2}\cos\theta)^2$ is on the circle $x^2+y^2=3$.

Setting $\theta=0$ we find $(1,-\sqrt{2})$, an example of an integer $2$-rational point.

Using $\cos\theta=\dfrac{4}{5},\,\sin\theta=\dfrac{3}{5}$ (as before) we find the non-integer $2$-rational point $\left(\frac{4}{5}+\frac{3}{5}\sqrt{2},\frac{3}{5}-\frac{4}{5}\sqrt{2}\right).$

Check: $\left(\dfrac{4}{5}+\dfrac{3}{5}\sqrt{2}\right)^2+\left(\dfrac{3}{5}-\dfrac{4}{5}\sqrt{2}\right)^2=3.$

1. Obtain a non-integer $2$-rational point on the circle $x^2+y^2=11$.

Note that $x^2 + y^2=11=1+10$ and we want a $\sqrt{2}$ rather than a $\sqrt{10}$, so some scaling is required.

Consider \begin{align*} (c\cos\theta&+d\sqrt{m}\sin\theta)^2+(c\sin\theta-d\sqrt{m}\cos\theta)^2\\ &=c^2\left(\cos^2\theta+\sin^2\theta\right) + d^2m\left(\cos^2\theta+\sin^2\theta\right)=c^2+d^2m. \end{align*} Setting $m=2$ again to get the square root correct, we need (integer) $c$ and $d$ such that $11=c^2+2d^2=3^2+2\times 1^2,$ so $c=3$ and $d=1$. Thus \begin{align*} p&=3\cos\theta \qquad r=3\sin\theta \\ q&=\sin\theta \qquad s=-\cos\theta \end{align*}

so taking the same values for $\cos\theta$ and $\sin\theta$ as before we find an example of a non-integer 2-rational point on the circle $x^2+y^2=3$: $\left(\frac{12}{5}+\frac{3}{5}\sqrt{2},\frac{9}{5}-\frac{4}{5}\sqrt{2}\right).$

1. Obtain a non-integer $2$-rational point on the hyperbola $x^2-y^2=7$.

There are few different ways to approach this. Here are three. Note that in all three we have a certain amount of freedom in how we choose our numbers.

Approach 1—Simple algebra

We know that if $(p+\sqrt{2}q, r+\sqrt{2s})$ lies on the hyperbola $x^2 - y^2 = 7$, then $p^2 + 2q^2 - r^2 -2s^2 + 2\sqrt{2}(pq - rs)=7.$

Thus we could sensibly choose $p^2 + 2q^2 - r^2 - 2s^2=7, \quad pq - rs=0.$

The second equation here suggests putting $p =ks, r = kq$, whereupon the first equation becomes $(k^2-2)(s^2-q^2) = 7$.

Now $k = 3$ is the obvious choice, giving $s^2 - q^2 = 1$, which yields $(s+q)(s-q) = 1$.

Putting $s+q = 2$ and $s-q = \dfrac{1}{2}$ gives $s = \dfrac{5}{4}, q = \dfrac{3}{4}$, and thus $p = \dfrac{15}{4}, r = \dfrac{9}{4}$.

Thus $\left(\dfrac{15}{4} + \sqrt{2}\dfrac{3}{4}, \dfrac{9}{4}+ \sqrt{2}\dfrac{5}{4}\right)$ is our non-integer 2-rational point, as required.

Approach 2—Trigonometry

Instead of the identity $\cos^2 \theta+\sin^2\theta=1$, we instead use $\sec^2\theta-\tan^2\theta=1$. We therefore try \begin{align*} x&=e\sec\theta+f\sqrt{m}\tan\theta\\ y&=e\tan\theta+f\sqrt{m}\sec\theta.\\ \end{align*}

Then $x^2-y^2=e^2-f^2m.$ As before, $m=2$ so we can make $7$ with the integers $e=3$ and $f=1$. Then $(x,y)=(3\sec\theta+\sqrt{2}\tan\theta,3\tan\theta+\sqrt{2}\sec\theta).$ Using $\cos\theta=\dfrac{4}{5}$ and $\sin\theta=\dfrac{3}{5}$ again, we find $\sec\theta=\dfrac{5}{4}$ and $\tan\theta=\dfrac{3}{4}$ and so we have $(x,y)=\left(\frac{15}{4} + \frac{3}{4}\sqrt{2}, \frac{9}{4}+\frac{5}{4}\sqrt{2}\right).$

Approach 3—Hyperbolic functions

We can also approach this problem using the hyperbolic functions, which makes sense as $x^2-y^2=7$ is a hyperbola. Note the identity $\cosh^2\theta-\sinh^2\theta=1,$ so try \begin{align*} x&=e\cosh\theta+f\sqrt{m}\sinh\theta\\ y&=e\sinh\theta+f\sqrt{m}\cosh\theta.\\ \end{align*} Then $x^2-y^2=e^2-f^2m.$ As before, $m=2$ so we can make $7$ with the integers $e=3$ and $f=1$. Then $(x,y)=(3\cosh\theta+\sqrt{2}\sinh\theta,3\sinh\theta+\sqrt{2}\cosh\theta).$ Then exploiting the definitions of $\cosh$ and $\sinh$ we have $\cosh\theta=\frac{e^{\theta}+e^{-\theta}}{2} \qquad \qquad \sinh\theta=\frac{e^{\theta}-e^{-\theta}}{2}.$ We can take $\theta=\log 2$ to find an example of a non-integer $2$-rational point: \begin{align*} \cosh(\log 2)&=\frac{e^{\log 2}+e^{-\log 2}}{2}=\frac{2+\frac{1}{2}}{2}=\frac{5}{4}\\ \sinh(\log 2)&=\frac{2-\frac{1}{2}}{2}=\frac{3}{4} \end{align*}

and so we obtain $(x,y)=\left(\frac{15}{4} + \frac{3}{4}\sqrt{2},\frac{9}{4}+\frac{5}{4}\sqrt{2}\right).$