Arithmetic Help

[FriendlyAsian]

Find positive integers x, y satisfying both x³ + y and y³ + x are divisible by x² + y²

 

[Romsek]

This is pretty straightforward. Do the divisions. Set the remainders to 0 and solve.

Is there anything in particular you are having trouble with?

 

[FriendlyAsian]

How can I do the divisions?

 

[FriendlyAsian]

I mean, I tried to do the divisons but I still can not solve it.

 

[Romsek]

\(\displaystyle x^3 + y = x(x^2+y^2) + (y-xy^2)\)
\(\displaystyle y^3 + x = y(x^2+y^2) + (x-x^2y)\)

The remainders are the far right terms on each line. They both need to be zero. I leave you to solve that.

 

[FriendlyAsian]

Thanks

 

[FriendlyAsian]

But what if both xy²- y and yx²- x are divisible by divisible by x²+ y²? In that case, these don't need to be zero.

 

[FriendlyAsian]

I've just figured out a solution. Can you guys check it for me
Since both xy² - y and yx² - x are divisible by x² + y², we get
(x + y)(x² + y²) - (x + y)(xy - 1) is divisible by x² + y²
=> (x+ y)(xy - 1) is divisible by x² + y²
Let x + y be a, xy be b (a, b are positive integers)
=> a(b - 1) is divisible by a² - 2b
=>a²b - a² is divisible by a² - 2b
We also have a² - 2b² is divisible by a² - 2b
=>a - 2b² is divisible by a² - 2b
Because a and b are positive integers so that 2b² [MATH]\ge [/MATH] 2b
=>a² - 2b [MATH]\ge [/MATH] a² - 2b². Since a - 2b² is divisible by a² - 2b
=>a - 2b² = a²
=>2b² = 2b
=>b = 1
=>xy = 1
=>x=y=1