My colleagues marked the first round of the British Mathematical Olympiad in Sidney Sussex College, Cambridge. You can view the leaderboard on Joseph’s website; well done to everyone who featured. There are quite a few new names on that list, which is rather promising since I may acquire actual mentees for the Advanced Mentoring Scheme.
Speaking of the AMS, one of the recent problems was a submission of mine: determine whether there exist solutions in positive integers to the Diophantine equation x^4 + y^6 = z^10. I understand that there is a non-empty intersection between people on that scheme and readers of cp4space, so I won’t reveal any answers here.
Instead, I’ll just talk about Diophantine equations of this form. If you only have three terms and the exponents have a common factor greater than 2, then no solutions exist due to Fermat’s last theorem. If the exponents are pairwise coprime, then it’s always possible to find solutions. (4,6,10) are neither of these, so I’ve not given away the solution to my problem!
My favourite Diophantine equation is x^4 + y^4 + z^4 = w^4. Essentially, this asks whether there exists a cuboid (rectangular parallelepiped) such that all three edge lengths and the triagonal are perfect squares. The smallest solution, discovered by Roger Frye using an exhaustive computer search, is 95800^4 + 217519^4 + 414560^4 = 422481^4.
As an exercise to the reader, you may want to prove that x^(4a) + y^(4b) + z^(4c) = w^(4d) also has solutions, where a,b,c,d are pairwise coprime integers.