# Find rational points on $x^2 + y^2 = 3$ and on $x^2 + y^2 = 17$

$(a)$ Find all rational points on the circle $x^2 + y^2 = 3$, if there are any. If there is none, prove so. $(b)$ Find all rational points on the circle $x^2 + y^2 = 17$, if there are any. If there is none, prove so.

I'm not sure how proceed with finding a general formula (if there is one)

I know that for $(a)$ there is no rational points but I don't know how to explain that there are none.

whereas for $(b)$ there are such points, $(1,4)$ for example. I think that we can find the intersection between the line $y=m(x-1)+4$ and $x^2 + y^2 = 17$

Any help is appreciated!

• Yes, you can get a parametric representation of the rational points on $x^2+y^2=17$ precisely using the line $y=m(x-1)+4$. Find the coordinates of the other meeting point. We can even (sort of) bypass solving a quadratic equation. Are you having trouble with details? – André Nicolas Mar 24 '15 at 1:17
• No, I think I can do that. I just needed some confirmation. Thanks though! – mike russel Mar 24 '15 at 1:27
• – nguyen quang do Feb 16 '19 at 16:45

$a)$ it amounts to solving in $\mathbb{Z}: x^2+y^2=3z^2$. You have that $x^2+y^2 = 0 \pmod 3 \to x = y = 0 \pmod 3$, and you get back the original one using descending method, and this proves $x = y = z = 0$, but this means the first equation $x^2+y^2 = 3$ has no rational solutions.

• Does this work for any prime $p$ (for equation $x^2+y^2=p$)? – lisyarus Mar 24 '15 at 1:16
• @lisyarus Only if $p\equiv 3\pmod{4}$. Otherwise, $x^2 + y^2 = p$ not only has nontrivial rational solutions, it has nontrivial integer solutions (e.g. $1^2 + 2^2 = 5$) – Slade Mar 24 '15 at 1:21
• @Slade got it, thanks. – lisyarus Mar 24 '15 at 1:37

For $N$ an integer, the general result is that if $x^2+y^2=N$ has rational solutions, then it has at least one integer solution.

As shown in this answer, $n$ can be written as the sum of two squares if and only if, in the prime factorization of $n$, each prime that is $\equiv3\pmod4$ appears with even exponent.

If $x^z+y^2=3z^2$, then $3$ appears with odd exponent. Thus, there are no rational solutions of $$\left(\frac xz\right)^2+\left(\frac yz\right)^2=3\tag{1}$$

As noted, $17=4^2+1^2$. Suppose that $$\left(\frac xz\right)^2+\left(\frac yz\right)^2=17\tag{2}$$ then \begin{align} 1 &=\frac{x^2+y^2}{17z^2}\\ &=\frac{x+iy}{z(4+i)}\frac{x-iy}{z(4-i)}\tag{3} \end{align} which means that \begin{align} \frac{x+iy}{z(4+i)}\tag{4} &=u+iv \end{align} where $u,v\in\mathbb{Q}$ so that $u^2+v^2=1$.

Thus, $$\frac xz+i\,\frac yz=(4+i)\left(\frac ac+i\,\frac bc\right)\tag{5}$$ where $a^2+b^2=c^2$ is a Pythagorean triple, all of which can be generated using the formula derived in this answer: \begin{align} a &= m^2 - n^2\\ b &= 2mn\\ c &= m^2 + n^2 \end{align}\tag{6} Using $(5)$ and $(6)$, we can compute all rational solutions of $(2)$.

Example

Using the Pythagorean triple $(3,4,5)$, we get $$\left(\frac35+i\,\frac45\right)(4+i)=\frac85+i\,\frac{19}5$$ and $$\left(\frac35-i\,\frac45\right)(4+i)=\frac{16}5-i\,\frac{13}5$$ Thus, we get $$\left(\frac85\right)^2+\left(\frac{19}5\right)^2=17$$ and $$\left(\frac{13}5\right)^2+\left(\frac{16}5\right)^2=17$$