This one comes down to contradiction. Think about what x, y, z can be. Show that if x, y, z are above a certain value, then the equation breaks down. That value may also be a variable.
I'm not sure if that helps, but it's best to compare how the ratios work if you just pick a number for each of x, y, z, then keep trying.
If it's still not working for you in a couple hours, message me and I will try to hint you again without giving up the answer.
This is a tough one to coach without just saying what it is. Sorry.
I thought about that, but the issue is that it seems like all the terms are sorta cancelling each other out. Like if x = y = z, then the left hand side is always equal to 1.5. On the other hand if we fix y and z and let x get large, then the left hand side grows very large. So it doesn't seem like there's an easy condition such that after a certain point the equation is impossible to be true.
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.
This one comes down to contradiction. Think about what x, y, z can be. Show that if x, y, z are above a certain value, then the equation breaks down. That value may also be a variable.
I'm not sure if that helps, but it's best to compare how the ratios work if you just pick a number for each of x, y, z, then keep trying.
If it's still not working for you in a couple hours, message me and I will try to hint you again without giving up the answer.
This is a tough one to coach without just saying what it is. Sorry.
I thought about that, but the issue is that it seems like all the terms are sorta cancelling each other out. Like if x = y = z, then the left hand side is always equal to 1.5. On the other hand if we fix y and z and let x get large, then the left hand side grows very large. So it doesn't seem like there's an easy condition such that after a certain point the equation is impossible to be true.
Think about what happens if x, y, z are all equal to 1....how would you make the denominator go away without that knowledge, though?
it may make sense just to brute force it by removing all denominators from the left side of the equation.
sometimes number theory means doing a bunch of stupid multiplications and then simplifying.
also, sometimes i purposefully make it painful, because discovering the solution is far more important, I am a Freirean educator...
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.