# Arithmetic Help

#### FriendlyAsian

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

#### Romsek

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

Is there anything in particular you are having trouble with?

• topsquark

#### FriendlyAsian

##### New member
How can I do the divisions?

#### FriendlyAsian

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

#### Romsek

##### Full Member
$$\displaystyle x^3 + y = x(x^2+y^2) + (y-xy^2)$$
$$\displaystyle y^3 + x = y(x^2+y^2) + (x-x^2y)$$

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

• • topsquark and Subhotosh Khan

Thanks

#### FriendlyAsian

##### New member
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

##### New member
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 $$\displaystyle \ge$$ 2b
=>a2 - 2b $$\displaystyle \ge$$ a2 - 2b2. Since a - 2b2 is divisible by a2 - 2b
=>a - 2b2 = a2
=>2b2 = 2b
=>b = 1
=>xy = 1
=>x=y=1