MHB How can $5^{1985}-1$ be factored into three integers greater than $5^{100}$?

[anemone]

Factor $5^{1985}-1$ into a product of three integers, each of which is greater than $5^{100}$.

 

[anemone]

Solution suggested by other:

Spoiler

Note that $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$ and $x^4+x^3+x^2+x+1=(x^2+3x+1)^2-5x(x+1)^2$, hence if we let $x=5^{397}$, we have

$\begin{align*}x^4+x^3+x^2+x+1&=(x^2+3x+1)^2-5x(x+1)^2\\&=(x^2+3x+1)^2-5^{398}(x+1)^2\\&=(x^2+3x+1)^2-(5^{199}(x+1))^2\\&=(x^2+3x+1+5^{199}(x+1))(x^2+3x+1-5^{199}(x+1)) \end{align*}$

It is obvious that $x-1$ and $x^2+3x+1+5^{199}(x+1)$ are both greater than $5^{100}$.

As for the third factor, we have

$x^2+3x+1-5^{199}(x+1)=x(x-5^{199})+3a-5^{199}+1 \ge a+0+1 \ge 5^{100}$

Hence $5^{1985}-1$ can be expressed as a product of three integers, i.e. $5^{1985}-1=(5^{397}-1)(5^{794}+3(5^{397})+1+5^{199}(5^{397}+1))(5^{794}+3(5^{397})+1-5^{199}(5^{397}+1))$, each of which factor is greater than $5^{100}$.