Number theory divisibility check question

N = $$2^{744} - 1$$. Prove N is divisible by $$2^{93}+2^{47}+1$$. I have no idea how to proceed. (edit: removed first part as I got the answer)

• observe that $$744/3=?$$ – lab bhattacharjee Apr 3 at 18:14
• @labbhattacharjee got the first one – Meet Shah Apr 3 at 18:17

2 Answers

If $$2^{31/2}=a,N=a^{48}-1$$

$$d=a^6+\sqrt2a^3+1$$

$$d$$ will divide

$$(a^6+1)^2-(\sqrt2a^3)^2=a^{12}+1$$

which again divides $$N$$

• I get that your answer is correct but how did you come up with the idea of writing $(a^6 +1) ^2 - (2^{1/2}a^3)^2$. – Meet Shah Apr 3 at 18:46
• @Meet, wanted to reach at $31$ as exponent – lab bhattacharjee Apr 3 at 18:56
• Thanks for helping! – Meet Shah Apr 3 at 19:04

By Aurifeuillean_factorization, $$2^{186}+1=(2^{93}+2^{47}+1)(2^{93}-2^{47}+1),$$

so $$(2^{93}+2^{47}+1)$$ divides $$2^{186}+1.$$

Then use $$n+1$$ divides $$n^4-1=(n+1)(n-1)(n^2+1)$$ with $$n=2^{186}$$ and you're done.