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My proof makes me think that the Fetmat's generalized equation is true.
No. The normal way is with 3 lines:
yes, but the normal way to write that is:
so x | y means y mod x = 0 , Thanks.
| means divides.
krassi, It appears you've been learning a lot of new stuff lately.
Here's my proof:
Let x=p+y, p is an integer. When we substitute in A and simplify, we get the equation:
This is very pretty. We'll take two cases:
1.p is even:
Then (p+2y) is even.
1.1.y is even
Then 2|3y and 2|p+y => 4|3y(p+y) => 4|p^2-3y(p+y) =>(p^2-3y(p+y))=+ or - 4. therefore we get that (p+2y) must be + or -1, but (p+2y) is even so y isn't even.
1.2. y is odd
Then 4 don't divides (p+2y) so p+2y is + or -2 and (p^2-3y(p+y))=+ of - 2.
But 2 doesn't divide 3y and 2 doesn't divide p+y so 2 doesn't divide 3y(p+y) so 2 doesn't divide p^2-3y(p+y) so p^2-3y(p+y) is odd but it must be +-2 so y is not odd.
2.p is odd:
p^2 is odd.
p+2y is odd too.
That means that p+2y=+-1 and p^2-3y(y+p) must be +-4
2.1. y is even:
then 3y is even and 3y(y+p) is even so p^2-3y(y+p) is odd which can't be +-4.
2.2. y is odd:
p is odd too so p+y is even so 3y(p+y) is even again and p^2-3y(p+y) is odd. And it just can't be +-4.
The proof is done.
Thanks, Diophantine Equations helped me, cause they led me to Fermat's Theorem(Generalized Equation: x^n+y^n=c*z^n)
well theres the sum and difference of two cubes identity, though I don't really see how that would help you here.
diophantine equations has something about integer answers.
"Want proof? Well, YOU try solving it!"