# 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

## 1 Answer

The simple answer is that Dipohantus solved all of his problems himself: There are no unsolved problems in his books.

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.

Of course the answer to the question whether all problems that are motivated by Diophantus have been solved is negative.