Show that the curve $x^2+y^2-3=0$ has no rational points

Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.

Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.

Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.

Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.

In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.

Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.

Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.

Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.

We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.

• By rational points do you mean a point where both the $x$ and the $y$ coordinates are rational numbers? – Pratyush Sarkar Mar 3 '13 at 16:30
• Yes, that exactly :) – ivan Mar 3 '13 at 16:31
• Write $x=p/q$, $y=r/s$ with $\gcd(p,q)=\gcd(r,s)=1$. The fact that the square of an odd number is congruent to $1\pmod8$ (or modulo 4) allows you to eliminate many cases. In a remaining case, you have to cancel some twos... – Jyrki Lahtonen Mar 3 '13 at 16:37
• The symbol $\mathbb{R}$ means the set of all real numbers. The symbol $\mathbb{Q}$ means the set of all rational numbers. I've changed your post accordingly. – Fly by Night Mar 3 '13 at 16:38
• I have a poem about this on my profile :) – Nick Jul 13 '18 at 19:52

Suppose to the contrary that there is a rational solution of the equation. Then there exist integers $a$, $b$, and $q$, with $q\ne 0$, such that $a^2+b^2=3q^2$, and $a$, $b$, and $q$ have no common factor greater than $1$.
Note that $a$ and $b$ must both be divisible by $3$. For if an integer $m$ is not divisible by $3$, then $m^2$ has remainder $1$ on division by $3$. So if one or both of $a$ and $b$ is not divisible by $3$, then $a^2+b^2$ has remainder $1$ or $2$ on division by $3$, and therefore cannot be of the shape $3q^2$.
Thus both $a$ and $b$ are divisible by $3$. It follows that $q$ is divisible by $3$, contradicting our assumption that $a$, $b$, and $q$ have no common divisor greater than $1$.
• I guess in your proof $x = a/q$ and $y = b/q$, but how do you know they will have the same denominator when the rational number is written in lowest terms? – Pratyush Sarkar Mar 3 '13 at 16:46
• @PratyushSarkar If you write both fractions with th elowest common denominator like this, then any two of $a,b,q4 may have a factor incommon, but not all three. – Hagen von Eitzen Mar 3 '13 at 16:49 • @HagenvonEitzen Ok. That makes sense. Thanks for the clarification. – Pratyush Sarkar Mar 3 '13 at 16:55 • Why does it follow that$q$is divisible by 3? – ivan Mar 3 '13 at 20:21 • We have$a$divisible by$3$, say$a=3c$. Similarly,$b=3d$. So$9c^2+9d^2=3q^2$. Therefore$3(c^2+d^2)=q^2$. So the prime$3$divides$q^2$, and therefore it divides$q$. – André Nicolas Mar 3 '13 at 20:25 Suppose$a^2 + b^2 = 3 c^2$Write$a = 3^p u$,$b = 3^q v$, and$c = 3^r w$, where$u, v, w$are all relatively prime to 3. This does not assume that$a, b, c$have no common divisor greater than 1. Assume$p \le q$(if$p > q$, switch their roles in what follows).$a^2+b^2 = (3^p u)^2 + (3^q v)^2 = 3^{2p}(u^2+ 3^{2(q-p)}v^2) $, so an even power of 3 divides$a^2+b^2$. ($u^2+ 3^{2(q-p)}v^2$has a remainder of 1 or 2 mod 3 depending on if$p < q$or$p = q$.) But$3 c^2 = 3 (3^r w)^2 = 3^{2r+1} w^2$, so an odd power of 3 divides$c^2$. By unique prime factorization, this is a contradiction. Note: I wrote this because the assumption of$a, b, c$having no common factor is, to me, either an implicit use of unique factorization or an infinite descent contradiction based on the powers of 3 dividing them. • I think you're onto something, and this seems like it might be what the book was going for. I'm not sure I follow it 100% yet, but my brain's kinda fried at this point. I'm gonna have to take another look at this when I'm fresh. – ivan Mar 4 '13 at 3:58 It is actually sufficient to analyze the curve modulo 3 (or 2, even), where the question becomes whether$-1 \equiv 2$is a square modulo 3. You can reduce the proof to this step directly by clearing denominators/divisibility arguments, but this curve is related to the quaternion algebra generated by the square roots of -1 and 3 which ramifies at the prime 3. Here, I'm making use of the Albert-Brauer-Noether-Hasse Theorem. Here is a more geometric flavored proof:$x^2+y^2-3=0 \iff x^2+y^2 = \sqrt{3}^2$is a circle with radius$\sqrt{3}$centered at the origin. Think of the points along the circle as polar coordinates$(r, \theta), i.e. \begin{equation}(\sqrt{3}, \theta) \text{ where } 0 \leq \theta \leq 2\pi \end{equation} The formulas for converting a polar coordinate into a cartesian coordinate is just right triangle trigonometry: \begin{align*} x &= r\cos\theta \\ y &= r\sin\theta \end{align*} Sincer=\sqrt{3}, we have that \begin{align*} x &= \sqrt{3}\cos\theta \\ y &= \sqrt{3}\sin\theta \end{align*} Then you can say that\sqrt{3}$multiplied by any number between$-1$and$1$is irrational. • Huh?$\sqrt{3}\sin\theta$crosses all rational points between$-\sqrt{3}$and$\sqrt{3}\$. – Fujoyaki Nov 6 '15 at 3:08