# 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
2 years, 6 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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 $$2^{42} + 2^{21} + 1 = 2^{36 + 6} + 2^{18+3} + 1 \equiv 2^6 + 2^3 + 1 = 64 + 8 + 1 = 73 \equiv 0 \pmod {73}$$

- 2 years, 6 months 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

- 2 years, 6 months 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 $$x = 2^{21}$$

- 2 years, 6 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!

- 2 years, 5 months ago