# Are problems in “Arithmetica” of Diophantus all solved now?

It's well-known that Diophantus had written ”Diophantus“ which contains many problems about solving arithmetic equations. I wonder whether all of them has been solved using modern techniques, as some of them are counting rational points on higher genus curves..

For example in problem $$17$$ of book $$VI$$, Diophantus poses a problem which comes down to finding positive rational solutions to $$y^2 = x^6 + x^2 + 1$$, which is of genus $$2$$, and it is solved using Chabauty-Coleman method as in Joseph Loebach Wetherell's PHD thesis in $$1998$$. Is there any more difficult and unsolved problem?

• The History of Science and Mathematics StackExchange may be a better place for this (interesting!) question. – Blue Jan 31 at 1:33
• math.stackexchange.com/questions/1401110/… – individ Jan 31 at 4:16
• Do you mean a special collection of problems about diophantine equations ? It is well known that no algorithm exists to solve every diophantine equation. So, it might be useful to show which equations you are referring to , maybe with a link to the collection. – Peter Jan 31 at 10:19
• @Peter For a very large class of equations. There are standard and General methods of solution. – individ Jan 31 at 17:26
• @individ This comment needs a clarification. Of course there are classes of equations that can easily be solved, and classes that are harder but still generally solveable. But "very large class" is quite broad. I suggest you give a survey over the classes that have been solved. I will gladly upvote it . – Peter Jan 31 at 19:26

The equation you mention does not show up in Diophantus. Diophantus has to solve the equation $$x^8 + x^4 + x^2 = z^2$$; setting $$z = x^4 + \frac14$$ immediately gives you $$x = \frac12$$, and then he is done.
Of course we can nowadays ask whether the equation has other rational solutions, and we might even want to determine them all. Setting $$z = xy$$ and cancelling $$x^2$$ we arrive at $$y^2 = x^6 + x^2 + 1$$. From a historical point of view, Whetherell has asked a question not in Diophantus (but motivated by one of his problems) and has given an answer.