The GCF of all "bad" numbers less than $2020^{2020}+1$

Question

A number is considered bad if it can be written in the form $2020^n+1$ for some positive integer $n.$ What is the greatest common divisor of all bad numbers less than $2020^{2020}+1?$

Can somebody please guide me on how to solve this?

Answer 1

So the sequence of 'bad' numbers looks like this:

$S=\{2020^1+1, 2020^2+1 \ldots 2020^{2019}+1\}$

Basically $S=2020^k$ for natural numbers $k$ from 1 to 2019

If you can prove that any two numbers in this sequence are coprime i.e that they don't share any factors then $\gcd (S)$ will have to be 1.

Let's try one that brings out nice factoring: $2020^m+1, 2020^{2m}+1$ for some choice of $m \in k$.

The second one can be written as:

$2020^{2m}-1+2=(2020^m+1)(2020^m-1)+2$

So if it was divided by $(2020^m+1)$ it would leave a remainder of $2$. If $2$ can divide $2020^m+1$, then it would be the greatest possible divisor between the two.

However $2020$ is even hence $2020^m+1$ is odd, making this situation an impossibility.

Hence these two can't share any factors and the final answer is $1$

Note: if this doesn't make sense you can try smaller corollaries like $4^2+1, 4^4+1$ and $3^2+1, 3^4+1$. The second leaves a possibility of 2 but the first one doesn't.