Recent content by tda1201

Re: triplets x^2 + y^2 =5z^3 Thank you both. Amazing how you think of this! I get it until your last sentence: 'That is always possible, since (in the inductive construction) it cannot happen that both ac+bd and ac−bd are multiples of 5 .' Can you please give me some explanation on this?

Thank you, I think this helps me a lot! You're right; I should start using LaTeX, but I'm still a bit shy at using it...

Let a, k , l , m e Z>1 and let a^k=1 (mod m) and a^l= 1 (mod m). Let d=gcd(k,l) Prove that a^d=1 (mod m). I get already confused at the start: Is it true that k|phi(m) (Lagrange) but k can also be a multiple of the order of a (mod m) and then it can be the other way round. Can anybody clarify...

Re: triplets x^2 + y^2 =5z^3 Ok, thank you. I tried so many triples but of course missed this one:o But, I like Serena, if you multiply both x0 and yo with the same factor a^k, then x and y aren't relatively prime anymore?

Re: triplets x^2 + y^2 =5z^3 I stumble upon the very first hint; I cannot even find one solution!.. Usually (x,y) isn’t 1 then… Am I missing something simple?

Re: triplets x^2 + y^2 =5z^3 O sorry; my mistake! The title is right! x^2 + y^2 =5z^3..?.

Hello! Could anyone give me a hint to solve this diophantine equation? Show how to construct an infinite amount of triples x, y and z e N with (x,y)=1 and x^2 + y^2=5z^2?

How can I find a primitive root modulo 169? I found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I’m sure there’s a smarter way than trying 2^the orders that divide...