Find all positive integers n such that n3−18n2+115n−391 is a perfect cube.
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Note that (∏d∣nd)2=∏d∣nd∏d∣nnd=nτ(n).
Therefore, we seek n such that nτ(n)=n6, that is, n=1 or τ(n)=6.
Write n=∏ppep. Then τ(n)=∏p(1+ep). There aren't many possibilities if this is to be 6 because each possibility corresponds to a factorisation of 6:
6=6 gives n=p5.
6=2⋅3 gives n=pq2.
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Therefore, we seek n such that nτ(n)=n6, that is, n=1 or τ(n)=6.
Write n=∏ppep. Then τ(n)=∏p(1+ep). There aren't many possibilities if this is to be 6 because each possibility corresponds to a factorisation of 6:
6=6 gives n=p5.
6=2⋅3 gives n=pq2.
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