Surprisingly, I managed to prove the following result:
For any rational , there exists an irrational such that is rational.

Suppose . consider the equation . We aim at proving that there is no rational satisfies this equation.

Assume in contrary that such an exists.
(1) : This gives , i.e., , violating our assumption.

(2) : Write for some positive coprime integers and .

Since is rational, is also rational. Therefore for some positive coprime integers and with .
Moreover, implies . So .
Substituting these formulae into the original equation, we get , or .
Since is irreducible and is also irreducible, we have and .
For , i.e., , since , we have or .
For , we have ; when , we have , which has or as its integer solutions.
(i) . Since , such does not exist. There is no solution.
(ii) . Put into the equation, and get , which does not have integer solution.
(iii) , . We have .
Try : , which is impossible. Try : is, again, impossible. So there is no solution either.
In conclusion, there is no rational satisfies the equation.

(3) : Write , where and are positive coprime integers. Again, for the same reason.

Plug in all these substitution to the equation: , or .
Again, by irreducibility, we have and .
(i) . Then and .
Note that . So . So , which is impossible.
(ii) . Then . Hence , which is impossible.
In conclusion, there is no rational number satisfies the original equation.

Now suppose . Consider . If there exists a rational satisfying this equation, then satisfies , contradicts to the above conclusion.

Now we prove that, for any rational , is rational for some irrational .

For , since , there exists with . It is proved that cannot be rational. So must be irrational.
For , we have . So there exists with . Again, cannot be rational by previous arguments. So must be irrational.
For , it is trivial that , for which is irrational but is rational.

This ‘proof’ is too clumsy and perhaps there are lots of errors. Please help me check if the arguments are valid.

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