How to prove that $10^{3n}+1$ is not a prime number?

Question

Show that the number 100...01, that include $3n-1$ zeros ($n\in N$) is not a prime number.

I found out that I could say $10^{3n}+1$ is the same as the requested proof but I don't know how to continue.

https://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd — lab bhattacharjee, Nov 07 '17 at 16:14

Austin Weaver · Accepted Answer · 2017-11-07T19:12:11.807

9

As you probably know, you simply need to factor the number in question to show its composite-ness.

In this case, there is a nice little factoring formula for two cubic numbers added together: $$a^3+b^3=(a + b)(a^2 - ab + b^2)$$ Thus: $$10^{3n}+1=(10^n+1)(10^{2n}-10^n + 1).$$ Therefore, both $10^n+1$ and $10^{2n}-10^n+1$ divide $10^{3n}+1$.

Additionally we must show that these two numbers are not $1$ or $10^{3n}+1$ itself, but I will leave this task to you.

edited Nov 07 '17 at 19:12

answered Nov 07 '17 at 15:33

Austin Weaver

2,059

1

Simple and effective. +1 – Cornman Nov 07 '17 at 15:38
Clean! +1 $, $ – Zubin Mukerjee Nov 07 '17 at 18:56

score 3 · Answer 2 · answered Nov 07 '17 at 15:36

3

Express the number as $(10^n)^3 + 1^3$. Now the formula for $a^3+b^3$ gives a factorization showing that this number is not a prime number (of course you have to check none the terms in the factorization is $1$).

answered Nov 07 '17 at 15:36

P Vanchinathan

19,504

score 3 · Answer 3 · answered Nov 07 '17 at 15:37

For odd $n$ it is easy to see, that $11|10^{3n}+1$ because of the alternating checksum, which is equal to $0$.

For even $n$ it is n=2m for natural $m\geq 1$ and we get:

$10^{6m}+1=10^{6m}-10^{4m}+10^{2m}+10^{4m}-10^{2m}+1=(10^{2m}+1)(10^{4m}-10^{2m}+1)$

How to prove that $10^{3n}+1$ is not a prime number?

3 Answers3