<s>[INST] 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?
 [/INST] 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.
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