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Factorise the number $5^{2015} - 1$ into three positive factors such that each is greater than $5^{200}$

I don't really know how to do this. Is inequality involved? Bonus question: can the same be done for $5^{2017} - 1$

user346756
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  • I could post a variant of this, but that would make this a duplicate :-). 2015 has a friendlier factorization than 1985, so it seems to me (see lhf's answer) that cyclotomic tricks will suffice. OTOH, if you want three factors $>5^{400}$, then Aurifeuillian is the way to go. And, no, I don't see a way to use Aurifeuillian factors with $5^{2017}-1$. – Jyrki Lahtonen Apr 06 '17 at 11:30

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Look at the factorization of $x^{2015}-1$ into cyclotomic polynomials: $$ x^{2015}-1 = \Phi_{1} \Phi_{5} \Phi_{13} \Phi_{31} \Phi_{65} \Phi_{155} \Phi_{403} \Phi_{2015} $$ and write $$ 5^{2015} - 1 = ABC $$ with $$ A=(\Phi_{1} \Phi_{5} \Phi_{13} \Phi_{31} \Phi_{65} \Phi_{155})(5), \quad B=\Phi_{403}(5), \quad C=\Phi_{2015}(5) $$ You still need to check that $A,B,C > 5^{200}$. The degrees of the polynomials are $215, 360, 1440$, which is a good indication (but not proof) that this is true.

You cannot do this trick for $2017$ because $2017$ is prime and $$ x^{2017}-1 = \Phi_{1} \Phi_{2017} $$ (but of course this does not mean that $5^{2017}-1$ cannot be factored)

lhf
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  • See also http://math.stackexchange.com/questions/2221051/growth-of-cyclotomic-polynomials. – lhf Apr 06 '17 at 16:04