Perfect Cube

For discussing Olympiad Level Number Theory problems
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Kazi_Zareer
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Perfect Cube

Unread post by Kazi_Zareer » Thu Aug 18, 2016 11:40 pm

If $ \frac{a }{b}+ \frac{b}{c}+ \frac{c}{a}$ is integer.
Show that $ abc$ is perfect cube.
We cannot solve our problems with the same thinking we used when we create them.

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Soumitra Das
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Re: Perfect Cube

Unread post by Soumitra Das » Wed Apr 05, 2017 7:51 pm

Here I want to use a simple lemma,which is:
Let a,b,c,d be some integer s.t. (a,b)=(c,d)=1.When $$\frac{a}{b}+\frac{c}{d}=n$$,where n \in N,then |b|=|d|
It's proof is simple,first prove b|d and second d|b
Proof of the original problem:
First,assume (a,b,c)=1.Let (a,b)=x,(b,c)=y,(c,a)=z.
$$\frac{a}{b}+\frac{b}{c}+\frac{c,a}=\frac{a/x}{b/x}+\frac{\frac{ab+c^2}{(c/y,a/z)yz}}{\frac{ac}{(c/y,a/z)yz}}$$ is an integer \Longrightarrow $$|b/x|=|\frac{ac}{(c/y,a/z)yz}|$$ [from the lemma]
But (y,a)=1.So,$$|b/x|=\frac{ac}{(c,a/z)yz}$$
By symmetry,$$|c/y|=\frac{ab}{(a,b/x)xz}$$ and |a/z|=\frac{bc}{(b,c/y)xy}.
Multiplying them,we get (c,a/z)(a,b/x)(b,c/y)=(a/z)(b/x)(c/y),which imply a/z|c ,b/x|a and c/y|b imply a|z^2 ,b|x^2 and c|y^2.Since x|a,x|z^2,which imply x=1 [since (a,b,c)=1]. So,|b|=1.Using similar argument,|a|=|c|=1.
Using it,when (a,b,c)=d,then |a|=|b|=|c|=d.Thus $$abc=\pm d^3$$,which is a perfect cube number.

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