]]>

x ≡ 0 (mod y)]]>

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.

]]>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).

]]>That's terrific.

What does the "|" symbol mean in above work?]]>

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.]]>

x³+y³=4(x²y+xy²+1) |*2

2x³+2y³= (x²y+xy²+1)*8[or: 2^3]

Therefore no integer-solutions are possible.

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

]]>can anyone give me a hint how to prove that the following equation can not be solved with integers for x and y?

x³+y³=4(x²y+xy²+1)

thanks

]]>