Decompose $5^{1985}-1$ into factors

Question

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?

what made you come to this question? is it an excercise from a book? — Max, Jul 20 '16 at 20:55
One factor is $5^{397}-1$. But according to Wolfram Alpha, The remaining factors are $2^2$, $11$, $71$, $146891$, $1043801929$ and $7768438039$ which unlikely fulfill all your requirements. — Ng Chung Tak, Jul 20 '16 at 21:01
@NgChungTak According to Pari-GP, $(5^{397}-1)/4$ is not prime, so other factors can come from this number — egreg, Jul 20 '16 at 21:05
@egreg Yes, it can further factorize. The factors $11$, $71$ and $146891$ seem belong to $\frac{5^{1985}-1}{5^{397}-1}$ — Ng Chung Tak, Jul 20 '16 at 21:05

Jyrki Lahtonen · Accepted Answer · 2016-07-20T21:43:18.977

18

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.

edited Jul 20 '16 at 21:43

answered Jul 20 '16 at 21:39

Jyrki Lahtonen

133,153

I thought of Aurifeuillian factorization, but did not expect that this works here. Great solution! – Peter Jul 20 '16 at 21:42
I thought Aurifeullian not come from a proper name, did not know that is because of J.F.A. Aurifeuille which I hear for the first time. Thanks. – Piquito Jul 20 '16 at 23:39
Nice answer, nice reference! (+1) – Markus Scheuer Jul 21 '16 at 07:05

score 3 · Answer 2 · answered Jul 20 '16 at 21:25

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.

score 3 · Answer 3 · edited Apr 13 '17 at 12:19

3

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 $$ 5^{5\cdot 397} = (5^{397}-1)\cdot\Phi_5\left(5^{397}\right).\tag{3} $$

edited Apr 13 '17 at 12:19

Community

1

answered Jul 20 '16 at 23:43

Jack D'Aurizio

353,855

1

How do you go from $$ \left(x^2-\frac{x}{2}+1\right)^2-\frac{5}{4}x^2$$ to $$(x^2+3x+1)^2-5x(x+1)^2?$$ – user19405892 Jul 21 '16 at 00:02
Inspired guess. – Jack D'Aurizio Jul 21 '16 at 00:03
Good job. Unless I'm mistaken, substituting $x=5t^2$ into $(2)$ leads to a difference of squares and exactly the Aurifeuillian factorization. – Jyrki Lahtonen Jul 21 '16 at 05:59

Decompose $5^{1985}-1$ into factors

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