Finding all rational points in $x^2+y^2=6$. I want to find all rational points in the circle $x^2+y^2=6$. This would be easy if I could find one rational point in the circle, however, it's very hard to guess one in this case. However, I don't think that this circle has no rational points and I wouldn't know how to try and prove this. 
 A: The  equation $$x^2+y^2=6z^2$$ has no non-zero integer solutions.
To see that, work $\pmod 3$.  The equation becomes $$x^2\equiv -y^2\pmod 3$$ which has no non-zero solutions (by inspection). But if $x\equiv y\equiv 0 \pmod 3$ then $z\equiv 0\pmod 3$ so a standard argument by infinite descent shows that there are no solutions.
This implies that there are no rational solutions to your equation.
A: Perhaps a little gaussian arithmetic could help. Consider the homogeneous diophantine equation $x^2+y^2=6z^2$ as written in $\mathbf Z [i]$ under the form $(x+iy)(x-iy)=3. (1+i)(1-i)z^2$, where $3$ is known to be inert (i.e. remains prime) and $2$ splits as the product of the two conjugate primes $(1+i)$ and $(1-i)$. The gaussian ring being a PID, the previous decomposition is unique up to units, which are $\pm 1 , \pm i$. The LHS being a product of two conjugate non associated quantities, the total exponent of $3$ in the RHS must be even : impossible.
This approach could serve in other analogous "linked" or "related" problems. 
