Arithmetic Help

FriendlyAsian

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Mar 22, 2019
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Find positive integers x, y satisfying both x3 + y and y3 + x are divisible by x2 + y2
 

Romsek

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Nov 16, 2013
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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

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Mar 22, 2019
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How can I do the divisions?
 

FriendlyAsian

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Mar 22, 2019
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I mean, I tried to do the divisons but I still can not solve it.
 

Romsek

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

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

FriendlyAsian

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Thanks
 

FriendlyAsian

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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

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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
 
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