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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)

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

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

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