The number $2021$ is fantabulous. For any positive integer $m$, if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous, then all the elements are fantabulous. Does it follow that the number $2021^{2021}$ is fantabulous?
Re: EGMO 2021 P1
Posted: Thu Apr 15, 2021 10:50 pm
by Mehrab4226
At first note,
$2021 \equiv -1\text{ (Mod 3)}$
$\therefore 2021^{2021} \equiv -1 \text{ (Mod 3)}$
So the number should change into $-1$ mod 3. So there are 3 cases
In $\text{mod 3}$
$m \equiv 0$
$2m+1 \equiv 1$
$3m \equiv 0$
$m \equiv 1$
$2m+1 \equiv 0$
$3m \equiv 0$
$m \equiv -1$
$2m+1 \equiv -1$
$3m \equiv 0$
So now we can simplify our problem to just focus on $\{m,2m+1\}$ where both elements must be $-1$ in mod $3$.
Now we can get a sequence,
$m,2m+1,2(2m+1)+1,\cdots$
If $2021^{2021}$ is a fantabulous number it must be in the sequence where $m=2021$.
But in $\text{Mod 4}$ we get,
$2021 \equiv 1$
$2021^{2021} \equiv 1$
In the sequence,
$m \equiv 1$
$2m+1 \equiv 3 \equiv -1$
$2(2m+1)+1 \equiv -1 $
So excluding the 1st term, all the other terms of the sequence are congruent to $-1 \text{ (Mod 4)}$ But since $2021^{2021} \equiv 1 \text{ (Mod 4)}$. $2021^{2021}$ is not in the sequence. So it is not a fantabulous number.
At first note,
$2021 \equiv -1\text{ (Mod 3)}$
$\therefore 2021^{2021} \equiv -1 \text{ (Mod 3)}$
So the number should change into $-1$ mod 3. So there are 3 cases
In $\text{mod 3}$
$m \equiv 0$
$2m+1 \equiv 1$
$3m \equiv 0$
$m \equiv 1$
$2m+1 \equiv 0$
$3m \equiv 0$
$m \equiv -1$
$2m+1 \equiv -1$
$3m \equiv 0$
So now we can simplify our problem to just focus on $\{m,2m+1\}$ where both elements must be $-1$ in mod $3$.
Now we can get a sequence,
$m,2m+1,2(2m+1)+1,\cdots$
If $2021^{2021}$ is a fantabulous number it must be in the sequence where $m=2021$.
But in $\text{Mod 4}$ we get,
$2021 \equiv 1$
$2021^{2021} \equiv 1$
In the sequence,
$m \equiv 1$
$2m+1 \equiv 3 \equiv -1$
$2(2m+1)+1 \equiv -1 $
So excluding the 1st term, all the other terms of the sequence are congruent to $-1 \text{ (Mod 4)}$ But since $2021^{2021} \equiv 1 \text{ (Mod 4)}$. $2021^{2021}$ is not in the sequence. So it is not a fantabulous number.
Recheck your proof $2021^{2021}$ IS FANTABULOUS.
HINTS
Every number is fantabulous and collatz conjecture(not necessary but if you know you will catch it quickly).
At first note,
$2021 \equiv -1\text{ (Mod 3)}$
$\therefore 2021^{2021} \equiv -1 \text{ (Mod 3)}$
So the number should change into $-1$ mod 3. So there are 3 cases
In $\text{mod 3}$
$m \equiv 0$
$2m+1 \equiv 1$
$3m \equiv 0$
$m \equiv 1$
$2m+1 \equiv 0$
$3m \equiv 0$
$m \equiv -1$
$2m+1 \equiv -1$
$3m \equiv 0$
So now we can simplify our problem to just focus on $\{m,2m+1\}$ where both elements must be $-1$ in mod $3$.
Now we can get a sequence,
$m,2m+1,2(2m+1)+1,\cdots$
If $2021^{2021}$ is a fantabulous number it must be in the sequence where $m=2021$.
But in $\text{Mod 4}$ we get,
$2021 \equiv 1$
$2021^{2021} \equiv 1$
In the sequence,
$m \equiv 1$
$2m+1 \equiv 3 \equiv -1$
$2(2m+1)+1 \equiv -1 $
So excluding the 1st term, all the other terms of the sequence are congruent to $-1 \text{ (Mod 4)}$ But since $2021^{2021} \equiv 1 \text{ (Mod 4)}$. $2021^{2021}$ is not in the sequence. So it is not a fantabulous number.
Ok. There is a problem with the solution,
Note,
$m \to 3m \to 6m+1 \to 12m+3 \to 4m+1 \to 2m $
So if 2m is fantabulous so is m, and it is the same if 2m+1 is fantabulous.
Now,
$2021 \to 1010 \to 505 \to 252 \to 126 \to 63 \to 31 \to 15 \to 7 \to 3 \to 2 \to 1.$
So 1 is fantabulous.
If any number $k$ is not fantabulous, then we can decrease it to 1 and get 1 is not fantabulous which is not true. Contradiction. So all numbers are fantabulous.
$\therefore 2021^{2021}$ is fantabulous.