Yes. This is right. thank you very much for your comments. we know that the soution of the equation x^2+y^2=2z^4 is (u^2,u^2,u). Then we get (2r^2p^2-s^2q^2=2s^2p^2-s^2q^2 , i.e., r=s or t=1 which leads to the trivial solution t=r/s=1. Then the case 2,4 has only trivial solution.

Dear Alekseyev ; very thanks, note that by letting x=p/q, t=r/s in the equation, we get the relation $p^2s^4+p^2r^4-q^2s^4-q^2s^2r^2=0$ which is not equivalent with $p^2s^4+p^2r^4-2r^2s^2q^2-2s^4q^2=0$.(this is obtained after some simplification in the above relation.) !! Am I saying right?

thanks, but t=0,1,-1 are trivial solution.

thank you very much for your valuable comments. I think that there exist integers n,m such that the equation has nontrivial solutions. What do you think about this?

please explain more about the first case.of course I know that Fermat case has not solution... And what do you think about the second example?

I do not understand how the problem converts to the above problem. By letting t=s/r we get x^2(s^4+r^4)=t^4+s^2r^2. then?

thanks for your comments.how do you convert my problem to Fermat problem?

You may see last version of our paper in the raxiv next week,too.

Thank you for your comments;You are right.This was the first version of my paper..we try to solve this problem in future....

I am sorry that I have given wrong answer about the motivation of my question regarding the EC that had posted on the website. As a matter of fact I didn't give the appropriate answer to secure our main problem of Diophantine equation.

Dear Prof Peter Mueller. I appreciate your reply and your answer. thank you.. It is great. How did you find this answer? Please more explain about why this point is not torsion?