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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.

McNuggets666
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  • https://math.stackexchange.com/questions/2773097/how-to-find-all-rational-solutions-of-x2-3y2-7/2788381#2788381 – individ Feb 16 '19 at 16:50

2 Answers2

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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.

lulu
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  • so what you are saying is that $3|x^2$ implies $3|x$, same with $y^2$ and therefore the equation can be written as $(3^pu)^2+(3^qv)^2=3(2(3^rw))$ and one side of the equation will always have an even power of 3 and the other one an odd power of 3, right? – McNuggets666 Feb 15 '19 at 21:08
  • @McNuggets666 Yes, that's correct. – lulu Feb 15 '19 at 21:16
  • @McNuggets666 I take that back. It's not obvious that the order of $3$ on the left is even. It's better to argue by infinite descent. Since $3$ divides each of $x,y,z$ if you had a solution we could always produce a smaller one, a clear contradiction. – lulu Feb 15 '19 at 22:07
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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.

nguyen quang do
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