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$(a)$ 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!

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  • $\begingroup$ 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? $\endgroup$ – André Nicolas Mar 24 '15 at 1:17
  • $\begingroup$ No, I think I can do that. I just needed some confirmation. Thanks though! $\endgroup$ – mike russel Mar 24 '15 at 1:27
  • $\begingroup$ See also math.stackexchange.com/a/3115195/300700 $\endgroup$ – nguyen quang do Feb 16 at 16:45
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$a)$ 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.

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  • $\begingroup$ Does this work for any prime $p$ (for equation $x^2+y^2=p$)? $\endgroup$ – lisyarus Mar 24 '15 at 1:16
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    $\begingroup$ @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. $1^2 + 2^2 = 5$) $\endgroup$ – Slade Mar 24 '15 at 1:21
  • $\begingroup$ @Slade got it, thanks. $\endgroup$ – lisyarus Mar 24 '15 at 1:37
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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.

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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 $$ \left(\frac{13}5\right)^2+\left(\frac{16}5\right)^2=17 $$

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