China TST 1987, problem 5

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tanmoy
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China TST 1987, problem 5

Unread post by tanmoy » Thu Jan 28, 2016 1:43 pm

Find all positive integer $n$ such that the equation $x^{3}+y^{3}+z^{3}=nx^{2}y^{2}z^{2}$ has positive integer solutions.
"Questions we can't answer are far better than answers we can't question"

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Kazi_Zareer
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Re: China TST 1987, problem 5

Unread post by Kazi_Zareer » Wed Apr 20, 2016 1:25 pm

Let us assume that $ x \geq y \geq z $
thus $ 3x^3 \geq x^3 + y^3 + z^3 = nx^{2}y^{2}z^{2} .$
So, $ x \geq \frac {ny^2z^2} {3} $,Because, $ x^{3} + y^{3} + z^{3} = nx^{2}y^{2}z^{2} $
thus,$ x^{2} | y^{3} + z^{3}$ , so $ 2y^{3} \geq y^{3} + z^{3} \geq x^{3} \geq \frac {n^2y^4z^4} {9} $

If $ z > 1 $ then $ y \leq \frac{18} {16x^2} $ so
$ y = 1 $ but $ y \geq z $- contradiction.

If $ z = 1 $ then $ y^{3} + 1 \geq \frac {n^2y^2} {9} $, so $ 9 + \frac{9} {y^3} \geq x^{2} y $

If $ y = 1 $ then $ x^{3} + 2 = nx^{2} $ so the only solution is $ (x, y, z, n) = (1, 1, 1, 3) $

$ ( x^{2} | 2 \Rightarrow x = 1) $ ,so $ n = 3 $ Satisfies

If $ y > 1 $ then $ 10 > n^{2} y $ so,

a) $ n = 1 $ then for instance $ (x, y, z, n) = (3, 2, 1, 1) $ is a solution so $ n = 1 $ also satisfies.
b)$ n = 2 $ then $ y = 2 $ or $ y = 1 $ then there are no solution - (easy to check)
c) $ n = 3 $ and $ y = 1 $ that gives $ (1, 1, 1, 3) $ what we had before.
(q.e.d)
(rafal)
So, answer:$ n = 1, 3 $
We cannot solve our problems with the same thinking we used when we create them.

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