Rational round

The following solution came from Alan of Madras College, St Andrew's.

Substituting

$$x=\frac{2k}{k^2+1},y=\frac{k^2-1}{k^2+1}$$

we get

$$\begin{array}{rl}{x}^2+y^2& ={\left(\frac{2k}{k^2+1}\right)}^2+{\left(\frac{k^2-1}{k^2+1}\right)}^2\\ & =\frac{(2k{)}^2+(k^2-1{)}^2}{(k^2+1{)}^2}\\ & =\frac{4k^2+k^4-2k^2+1}{(k^2+1{)}^2}\\ & =\frac{(k^2+1{)}^2}{(k^2+1{)}^2}\\ & =1.\end{array}$$

So for every integer $k$ the point $(x,y)$ lies on the circle $x^2 + y^2 = 1$. (Note $(k^2 + 1) \neq 0$.)

Assume that it is possible for rational points to lie on the circle $x^2 +y^2 = 3$. For example $${\left(\frac{m}{n}\right)}^2+{\left(\frac{k}{l}\right)}^2=3$$ where $m$, $n$, $k$, $l$ are integers, $n$, $l \neq 0$, gcd $(m,n)$ = 1, gcd $(k,l)$ = 1. This can be rearranged to give $(ml)^2 + (kn)^2 = 3(nl)^2$ and hence

$$(ml{)}^2+(kn{)}^2\equiv 0(\text{mod }3).$$

For any integer $q$ the only possibilities are

$$\begin{array}{rl}q& \equiv 0(\text{mod }3)\Rightarrow q^2\equiv 0(\text{mod }3)\\ q& \equiv 1(\text{mod }3)\Rightarrow q^2\equiv 1(\text{mod }3)\\ q& \equiv 2(\text{mod }3)\Rightarrow q^2\equiv 4\equiv 1(\text{mod }3)\end{array}$$

So the only possible way for $(ml)^2 + (kn)^2 \equiv 0$ (mod 3) is for $(ml)^2 \equiv 0$ (mod 3) and $(kn)^2 \equiv 0$ (mod 3).

Case 1: suppose that $m \equiv 0$ (mod 3). Then $n$ not$\equiv 0$ (mod 3) because gcd $(m,n) = 1$, so that $k \equiv 0$ (mod 3). Then $l$ not $\equiv 0$ (mod 3) because gcd $(k,l)$ = 1. We write $m = 3m_1$ and $k = 3k_1$ where $m_1$ and $k_1$ are integers; then as $(ml)^2 + (kn)^2 = 3(nl)^2$ we have $$3(m_1^2{l}^2+k_1^2{n}^2)=(nl{)}^2.$$ This shows that $nl \equiv 0$ (mod 3) which is a contradiction.

Case 2: suppose that $m$ not$\equiv 0$ (mod 3). Then $l \equiv 0$ (mod 3) and by the same reasoning as before we reach a contradiction.

Our assumption must be wrong, therefore no rational points lie on the circle $x^2 +y^2 = 3$.