I really don't see what you're saying. Like 1/2+1/3+1/6 = 1, why can't something similar happen here?
Clearing the denominators and doing some algebra I get x3+y3+z3−5xyz-3(x2y+xy2+x2z+xz2+y2z+yz^2) = 0 . This doesn't exactly look easier... I tried a parity argument, but it doesn't work. I don't think anything like that can work since multiplied out like this we introduced (0, 0, 0) as a solution and I wrote a python program to brute force this and I found solutions such as (-1, 4, 11) if we allow for negative integers.
A lot of it here is just trying a bunch of what we called "numericals" in college. Just plugging in numbers for x, y, z and looking for the structure when we simplified.
Once you get to the expanded algebraic equation, which I'm pretty sure you've got right, you should be able to show that the solution isn't possible for positive integers.
What must be true about that equation for it all to equal zero? I see a couple negatives, it may make sense to move them to the other side of the equation to find a statement equating two cubic relations. Pay close attention to the structure of the equation when you work with it. What happens with the odd numbers, prime numbers, positive and negative signs, powers, etc.
Ok I have to come clean, this is a bit. There do exist integer solutions, but you have to use algebraic geometry and in particular the theory of elliptic curves to solve this. The smallest solution has about 80 digits for each of x, y, and z.
I really don't see what you're saying. Like 1/2+1/3+1/6 = 1, why can't something similar happen here?
Clearing the denominators and doing some algebra I get x3+y3+z3−5xyz-3(x2y+xy2+x2z+xz2+y2z+yz^2) = 0 . This doesn't exactly look easier... I tried a parity argument, but it doesn't work. I don't think anything like that can work since multiplied out like this we introduced (0, 0, 0) as a solution and I wrote a python program to brute force this and I found solutions such as (-1, 4, 11) if we allow for negative integers.
A lot of it here is just trying a bunch of what we called "numericals" in college. Just plugging in numbers for x, y, z and looking for the structure when we simplified.
Once you get to the expanded algebraic equation, which I'm pretty sure you've got right, you should be able to show that the solution isn't possible for positive integers.
What must be true about that equation for it all to equal zero? I see a couple negatives, it may make sense to move them to the other side of the equation to find a statement equating two cubic relations. Pay close attention to the structure of the equation when you work with it. What happens with the odd numbers, prime numbers, positive and negative signs, powers, etc.
Ok I have to come clean, this is a bit. There do exist integer solutions, but you have to use algebraic geometry and in particular the theory of elliptic curves to solve this. The smallest solution has about 80 digits for each of x, y, and z.
lol, next time i'll open the chain letter from grandma.
this is a good bit, haven't seen it before that i can remember.