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Factoring a what?

This is a question from the International Mathematics Contest 2013, I've been trying to solve it for a long time, But I couldn't find a legitimate solution. Could anyone give a mathematical solution to it? Algebraic, Geometric (if something of the kind exists), Thanks! =D

Give a factor of the expression: \[2^{42} + 2^{21} + 1\].

Note by Giwon Kim
8 months ago

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Seems easy for a IMC Question, unless I messed up.

\( 2^{42} + 2^{21} + 1 = \frac{2^{63} - 1}{2^{21} - 1} \)

So we need to find a prime for which \( 2^{63} \equiv 1 \pmod{p} \) but \( 2^{21} \not \equiv 1 \pmod{p} \).

Since \( 9|63 \) and \( 9\not | 21 \), if we find a prime such that \( 2^9 \equiv 1 \pmod{p} \) but \( 2^3 \not \equiv 1 \pmod{p} \), we are done.

Now \( 2^9 - 1 = 511 = 7*73 \). So \( 2^9 \equiv 1 \pmod{7,73} \). But, we have \( 2^3 \equiv 1 \pmod{7} \) and \( 2^3 \not \equiv 1 \pmod{73} \). So 7 is ruled out. So the prime which we require is \( 73 \).

We can verify this since \( 2^{42} + 2^{21} + 1 = 2^{36 + 6} + 2^{18+3} + 1 \equiv 2^6 + 2^3 + 1 = 64 + 8 + 1 = 73 \equiv 0 \pmod {73} \) Siddhartha Srivastava · 8 months ago

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@Siddhartha Srivastava Thank you so much for the response! =D But could I ask one more thing? I can see you used Fermat's little Theorem there, but I don't understand how you factored the first equation. I'd really appreciate it if you could explain that part too. Thaaaanks =DD

EDIT: Or wait, is it Fermat's little theorem??? XDD Giwon Kim · 7 months, 3 weeks ago

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@Giwon Kim

\( 2^{42} + 2^{21} + 1 = \frac{2^{63} - 1}{2^{21} - 1} \)

This equation?

It's just some simple algebra. \( x^2 + x + 1 = \dfrac{(x^2 + x + 1)(x-1)}{(x-1)} = \dfrac{x^3 - 1}{x-1} \), where \( x = 2^{21} \) Siddhartha Srivastava · 7 months, 3 weeks ago

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@Siddhartha Srivastava Ohhhh I see, XD The idea is quite simple, but I've never encountered it before. That really helped a lot, XDD Thanks! Giwon Kim · 7 months, 1 week ago

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