$x+y\mod 5$ is not zero so we will never have $x+y\mod 5 = 0$ i mean if you multiply $x+y$ in anything except 5 you will never have $x+y\mod 5 = 0$

Why mod 5 instead of something else?

These types of problems typically require you to find something that doesn't change (at least, in the ways you don't want it to). The first way of moving around means you want to combine $x$ and $y$ with a commutative operator. Take, for example, addition. The last two ways of moving around means this sum may change $\pm 5$, so we have to compensate by taking it $\mod{5}$. The rest of the changes are explained by the solutions above.

The sum of the $x$ and $y$ coordinates is always going to leave a remainder of $1$ when divided by $5$. Hence, it's not possible to travel to the origin.

Are. You. Joking. I spent way too long doing useless garbage here. Take $x+y$ mod $5$ and realize the first move does nothing, the second and third multiplies by $3,$ and the fourth and fifth do nothing.

Consider the sum of the coordinates modulo 5. If Mr. Fat moves to $(3x, -2y)$ or $(-2x, 3y)$, the sum of the coordinates will be $$3x-2y = 3(x+y) - 5y,$$or the symmetric equivalent. However, if $x+y$ was not originally a multiple of 5, the resulting sum can never be a multiple of 5. Similarly, if Mr. Fat moves to $(x+1, y+4)$ or the equivalent, his sum of coordinates modulo 5 does not change. However, he starts on a position with sum 1 mod 5. Therefore, it is impossible for Mr. Fat to end up in any position with sum of coordinates 0 mod 5, of which $(0, 0)$ is one.

$x+y \pmod{5}$ gives either nothing change or $\times 3 \implies \neq 0$

Misread as Mr. Fat wandering around the lettuce

We claim that $x+y\not\equiv0\pmod5$ for all points $(x,y)$ which Mr. Fat visits. It is true for $(x,y)=(0,1)$, now suppose that Mr. Fat is at $(a,b)$ with $a+b\not\equiv0\pmod5$ and moves to $(c,d)$. If $(c,d)=(b,a)$ then $c+d\equiv a+b\not\equiv0\pmod5$. If $(c,d)=(3a,-2b)$ then $c+d\equiv 3(a+b)\not\equiv0\pmod5$. If $(c,d)=(-2a,3b)$ then $c+d\equiv 3(a+b)\not\equiv0\pmod5$. If $(c,d)=(a+1,b+4)$ then $c+d\equiv a+b\not\equiv0\pmod5$. If $(c,d)=(a-1,b-4)$ then $c+d\equiv a+b\not\equiv0\pmod5$. We are done by induction.

v_Enhance wrote: Mr. Fat moves around on the lattice points according to the following rules: From point $(x,y)$ he may move to any of the points $(y,x)$, $(3x,-2y)$, $(-2x,3y)$, $(x+1,y+4)$ and $(x-1,y-4)$. Show that if he starts at $(0,1)$ he can never get to $(0,0)$. Mr. Fat??!! lolololololololololololololol

$\mod 5$ cases and done

For all possible points $(x,y)$, we have $5\nmid x+y$ so $(0,0)$ is impossible.

Note that $5\nmid x+y$ as the first, fourth, and fifth operations do nothing to the sum modulo $5$ and the second and third multiply the sum by $3.$ Hence, $(x,y)=(0,0)$ is impossible. $\square$

For any $x$,$y$, let $x+y \equiv k \mod 5$. It is easy to see that $k$ does not change (invariant) throughout all operations, so if $x+y \equiv 1 \mod 5$ then it is clear that $x+y \equiv 0 \mod 5$ is never achievable, proving the assertion. $\blacksquare$

We define the moves as $A, B, C, D$ accordingly. For the sake of contradiction, we assume that $(0, 0)$ is achievable. Claim -- If so, the last move was $D$. Proof. $A$ can't be the last move, trivial. $B$ and $C$ can't be the last move, because the last point reached before $(0, 0)$ would be the same, hence impossibility. Since $D$ was the last move, it means $(-1, -4) \to (0, 0)$. The 2nd last position was $(-1, -4)$. Claim --- At least one of them stays non-negative always. Proof. We prove it by induction. We set $n = 2$ to be the base case, for which the proof is trivial. Now, we assume $n = k$ is true, it's time for $k + 1$. \[A(n) \implies \text{Unchanged}\]\[B(n) \implies \text{At least one stays nonnegative}\]\[C(n) \implies \text{At least one stays nonnegative}\]\[D(n) \implies \text{Before the operation we have at least one nonnegative, hence we have at least one positive.}\]Hence, $(-1, -4)$ cannot be reached. Hence contradiction.

Note that $x+y \pmod 5$ either remains the same or is multiplied by $3$. Thus, if we start at $1$, it can never reach $0$.

Use an invariant mod 5 on $x+y$. We need $x+y\equiv 0\pmod 5$. Notice how the first operation doesn't change $x+y$, the second and third only multiply by 3 (for example, $3x-2y=3(x+y)-5y$, so all it does is multiply by 3), and the last 2 keep the residue the same. Therefore, it can never be $0\pmod 5$, so he cannot get to $(0, 0)$.

Note that $x+y \equiv 0\pmod{5},$ needs to happen. Now, all transformations, keep the residue the same besides the second and third, which multiply the residue by $3,$ so a power of $3,$ needs to be divisible by $5,$ but contradiction $\blacksquare$

Hmm, I do not see this invariant above. Note that $(x+y)^4$ remains invariant under this process. But $(0+1)^4 \equiv 1\pmod{5}$ and $(0+0)^4 \equiv 0 \pmod{5}$. So not possible.

Let $S = x + y \bmod{5}$ where $(x, y)$ is Mr. Fat's current position. It is clear that $S$ can only change to $S$ or $3S$ with each move, and so if $S$ is not zero then $S$ cannot be zero after any sequence of moves. This finishes the problem as $S = 1$ initially and we want to show $S \neq 0$.

Two words: $\pmod{5}$.

Note that $x + y \pmod{5}$ either stays the same or multiples by $3$ after every move. Originally it is $1$ so it can never be $0$, so Mr.Fat can never reach $(0, 0)$.

Taking $mod \ 5$, we see that the sum of coordinates remains the same or gets multiplied by $3$. Since we start with the sum being $1(mod \ 5)$, we cannot reach $0(mod \ 5)$.

We will take $\pmod 5$. Call a point good if the coordinates sum to a number that is not divisible by $5$. Note that good points can only move to good points (one can verify this by looking at all the moves individually). Since we start at a good point, we can never move to $(0,0)$, a not-good point.