Decompose $5^{1985}-1$ into factors 
Decompose the number $5^{1985}-1$ into a product of three integers, each of which is larger than $5^{100}$.

We first notice the factorization $x^5-1 = (x-1)(x^4+x^3+x^2+x+1)$. Now to factorize $x^4+x^3+x^2+x+1$ we get $$(x^2+ax+1)(x^2+bx+1) = x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1 = x^4+x^3+x^2+x+1$$ implies $a+b = 1,ab+2 = 1$. Thus, $$x^4+x^3+x^2+x+1 = (x^2+\left(\frac{1+\sqrt{5}}{2}\right)x+1)(x^2+\left(\frac{1-\sqrt{5}}{2}\right)x+1).$$ Is it possible to continue from this approach because now the factors I have aren't integers or is there a better way?

A low-tech solution.
$1985=5\cdot 397$, and $ \Phi_5(x) $ is a palyndromic polynomial, decomposable as
$$ \left(x^2-\frac{x}{2}+1\right)^2-\frac{5}{4}x^2 \tag{1}$$
or as:
$$(x^2+3x+1)^2-5x(x+1)^2 \tag{2}$$
that with $x=5^{397}$ is the difference of two squares, $a^2-b^2=(a-b)(a+b)$, with both $a-b$ and $a+b$ being $>5^{100}$. The claim hence follows from
$$ 5^{5\cdot 397} = (5^{397}-1)\cdot\Phi_5\left(5^{397}\right).\tag{3} $$
