I don't have time to work on this, but you may find this thread helpful
http://www.mathhelpforum.com/math-he...ns-148115.html
And the proof technique of this post (#22) might apply
http://www.mathhelpforum.com/math-he...tml#post525708
I've read through a few exercises on Number Theory with regards to Diophantine Equations. Apparently, some of them are solvable and some are not. Here is the example that conflicts me:
Are there solutions to the Diophantine equation ? If so, what are they?
I don't have time to work on this, but you may find this thread helpful
http://www.mathhelpforum.com/math-he...ns-148115.html
And the proof technique of this post (#22) might apply
http://www.mathhelpforum.com/math-he...tml#post525708
undefined, I read that thread you had mentioned and the Wiki articles as well, however those threads had a lot of differing techniques and it was for a specific sample that involved the GCD of the variables equalling 1. I guess what I"m getting at was it looked like you, roninpro, and chip had many different approaches for the same problem, and I tried applying the proof technique of post #22 but didn't get anywhere.
If you or someone else does have a little time to spare, I would really like to see if there are any sols to this and if so find out what they are.
Much thanks for any help on this! - Brim
It's simpler to use Fermat's infinite descent:
Show that the only solution to is
Then suppose are solutions to the original problem. From above we get:
Therefore and we can write:
Therefore if is a solution then so is , and applying again and again:
is a soloution for all , and in particular made of integers.
But it's impossible that these numbers are integers for all if one of is nonzero, so we have the only solution:
We can generalize this problem:
Consider the diophantine equation with satisfying the conditions:
1) , the "squarefree part" of (that is, where is squarefree) is odd.
2) are coprime to .
3) (Jacobi Symbol)
then the only solution to the diophantine equation is the trivial solution.
Note that in the special case of , (2) is obviously true and we can replace condition (3) with:
(the question which started the thread is an obvious special case of the above)
Proof:
so it's enough to prove that has no nontrivial solutions.
Let be a solution. Then
.
Suppose that , and let be a prime number. Using the same idea of infinite descent as in my first post in this thread, we get a contradiction.
Thus and therefore:
which contradicts . Therefore . By symmetry, , and we can write:
.
(here we refer to the usual fractions, not Jacobi symbols)
Dividing by and using the fact that it's squarefree we get . Then
is another solution to the diophantine equation. Continuing ad infinitum yields a contradiction if not all are zero.
We thus get the desired result.