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Topic review (newest first)

krassi_holmz
2006-03-03 21:19:24

My proof makes me think that the Fetmat's generalized equation is true.
smile smile wink

krassi_holmz
2006-03-03 21:17:29

No. The normal way is with 3 lines:
x ≡ 0 (mod y)

Ricky
2006-03-03 14:06:32

yes, but the normal way to write that is:

y = 0 mod x.

Don't get me wrong, your way is perfectly fine too.  I guess I'm just used to seeing the mod and 0 on the right.

John E. Franklin
2006-03-03 08:08:03

so x | y  means  y mod x = 0 ,  Thanks.

Ricky
2006-03-03 08:04:16

| means divides.

2 | 4
3 | 30

The equivalent of:

x | y

is:

y = xk, where k is some integer

That is, y is a multiple of x (and k).

John E. Franklin
2006-03-03 05:47:39

krassi, It appears you've been learning a lot of new stuff lately.
That's terrific.
What does the "|" symbol mean in above work?

krassi_holmz
2006-03-03 01:21:37

Here's my proof:
Let A is the equation:


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.
4|p^2.
  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.

mling
2006-03-02 18:30:51

Thanks, Diophantine Equations helped me, cause they led me to Fermat's Theorem(Generalized Equation: x^n+y^n=c*z^n)

x+y=4(xy+xy+1)  |*2
2x+2y= (xy+xy+1)*8[or: 2^3]
Therefore no integer-solutions are possible.

mikau
2006-02-28 13:32:37

well theres the sum and difference of two cubes identity, though I don't really see how that would help you here.

John E. Franklin
2006-02-28 10:24:32

diophantine equations has something about integer answers.

ryos
2006-02-28 09:38:24

"Want proof? Well, YOU try solving it!"

I have nothing of value to add, just my lame sense of humor.

mling
2006-02-28 05:58:41

Hi,
can anyone give me a hint how to prove that the following equation can not be solved with integers for x and y?
x+y=4(xy+xy+1)

thanks

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