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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
11 months, 3 weeks 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 $$2^{42} + 2^{21} + 1 = 2^{36 + 6} + 2^{18+3} + 1 \equiv 2^6 + 2^3 + 1 = 64 + 8 + 1 = 73 \equiv 0 \pmod {73}$$ · 11 months, 3 weeks ago

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 · 11 months, 3 weeks ago

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

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 $$x = 2^{21}$$ · 11 months, 3 weeks ago