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USAMO 1973 #5 August 21, 2009

Posted by lumixedia in Uncategorized.
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The claim this problem makes looks obvious but turns out to be…actually, not much harder than it looks. The solution has one clever moment but I think you can avoid it by a little more algebra, though it’s not interesting enough to write out.

USAMO 1973 #5. Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.

Solution. Suppose otherwise. Then we have primes x<y<z so that

\displaystyle \frac{z^{1/3}-y^{1/3}}{y^{1/3}-x^{1/3}}=\frac{p}{q}

where {p} and {q} are positive integers and {(p,q)=1} (that is, in order for {x^{1/3}}, {y^{1/3}}, {z^{1/3}} to be terms of an arithmetic progression, the ratio of {z^{1/3}-y^{1/3}} to {y^{1/3}-x^{1/3}} should be rational since both differences are integer multiples of the common difference of the progression). Rearranging gives

\displaystyle qz^{1/3}+px^{1/3}=(p+q)y^{1/3}

\displaystyle q^3z+3q^2pz^{2/3}x^{1/3}+3qp^2z^{1/3}x^{2/3}+p^3x=(p+q)^3y

\displaystyle 3qpz^{1/3}x^{1/3}(qz^{1/3}+px^{1/3})=(p+q)^3y-q^3z-p^3x

\displaystyle 3qpz^{1/3}x^{1/3}(p+q)y^{1/3}=(p+q)^3y-q^3z-p^3x

\displaystyle (xyz)^{1/3}=\frac{(p+q)^3y-q^3z-p^3x}{3qp(p+q)}.

That is, {(xyz)^{1/3}} must be rational. Since {x,y,z} are distinct primes, this is clearly impossible. We conclude that there are no such {x,y,z}.

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Comments»

1. Qiaochu Yuan - August 21, 2009

One can make a much stronger claim; distinct roots of distinct primes must always be linearly independent. I sketched a proof for the square root case here and linked to an article by Iurie Boreico detailing an elementary proof of the general case. For cube roots one uses cubic reciprocity instead of quadratic reciprocity.


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