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: See here
http://factordb.com/index.php?query=5%5E1985-1
for the partial factorization of $5^{1985}-1$. There are three factors with $200$ digits or more. No idea how the factors can be expressed. Multiply one of the factors with the smaller factors to get a decomposition of the desired form.
A: 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
$$ 5^{5\cdot 397} = (5^{397}-1)\cdot\Phi_5\left(5^{397}\right).\tag{3} $$
A: I would try Aurifeuillian factorization here.
Let $f(x)=x^4+x^3+x^2+x+1$ be the fifth cyclotomic polynomial. The Aurifeuillian stuff says that $f(5x^2)$ factors into a product as follows
$$
f(5x^2)=\left(25 x^4-25 x^3+15 x^2-5 x+1\right) \left(25 x^4+25 x^3+15 x^2+5
   x+1\right).
$$
Call those two factors on the RHS $g_1(x)$ and $g_2(x)$.
Your number is
$$
\begin{aligned}
5^{1985}-1&=(5^{397}-1)f(5^{397})\\
&=(5^{397}-1)g_1(5^{198})g_2(5^{198}),
\end{aligned}
$$
and this fits the bill.
