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: \[2^{42} + 2^{21} + 1\].

## Comments

Sort by:

TopNewestSeems 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 · 1 year, 2 months ago

Log in to reply

EDIT: Or wait, is it Fermat's little theorem??? XDD – Giwon Kim · 1 year, 2 months ago

Log in to reply

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 · 1 year, 2 months ago

Log in to reply

– Giwon Kim · 1 year, 2 months ago

Ohhhh I see, XD The idea is quite simple, but I've never encountered it before. That really helped a lot, XDD Thanks!Log in to reply