Notation and definitions

Define ${\displaystyle \varphi ={\frac {{\sqrt {5}}-1}{2}}\approx 0.61803\,39887\ldots }$. (In other words, ${\displaystyle \varphi }$ here denotes the reciprocal of the golden ratio.) Observe that ${\displaystyle \varphi ^{2}=1-\varphi }$.

Let the target square be labeled with the value ${\displaystyle \varphi ^{0}=1}$, and all other squares be labeled with the value ${\displaystyle \varphi ^{n}}$, where ${\displaystyle n}$ is the Manhattan distance to the target square. Then we can compute the "score" of a configuration of soldiers by summing the values of the soldiers' squares. For example, a configuration of only two soldiers placed so as to reach the target square on the next jump would have score ${\displaystyle \varphi ^{1}+\varphi ^{2}}$.

When a soldier jumps over another soldier, there are three cases to consider:

1. When a soldier jumps towards the target square: Let the value of the soldier's square be ${\displaystyle \varphi ^{n}}$ for some ${\displaystyle n}$, and the value of the square he jumps over be ${\displaystyle \varphi ^{n-1}}$; then the total change in score after the jump is ${\displaystyle \varphi ^{n-2}-\varphi ^{n-1}-\varphi ^{n}=\varphi ^{n-2}(1-\varphi -\varphi ^{2})=0}$.
2. When a soldier remains the same distance from the target square after his jump: In this case the change in score is ${\displaystyle \varphi ^{n}-\varphi ^{n-1}-\varphi ^{n}=-\varphi ^{n-1}}$.
3. When a soldier jumps away from the target square: Here the change in score is ${\displaystyle \varphi ^{n+2}-\varphi ^{n+1}-\varphi ^{n}=\varphi ^{n}(\varphi ^{2}-\varphi -1)=-2\varphi ^{n+1}}$.

So, no jump will ever increase the configuration's total score.

Computing the score of the initial configuration

Consider now a starting configuration where only one infinite horizontal line is completely filled with soldiers.

If this horizontal line of soldiers is immediately below the target square, then the score of the configuration is ${\displaystyle \varphi +2\varphi ^{2}+2\varphi ^{3}+\ldots }$. The score of a line two spaces below the target square is ${\displaystyle \varphi ^{2}+2\varphi ^{3}+2\varphi ^{4}+\ldots =\varphi (\varphi +2\varphi ^{2}+2\varphi ^{3}+\ldots )}$. The score of a line three spaces below is ${\displaystyle \varphi ^{2}(\varphi +2\varphi ^{2}+2\varphi ^{3}+\ldots )}$, and so on.

Consider the full starting configuration, where soldiers fill the whole half-plane below the red line. This configuration's score is the sum of the scores of the individual lines. Therefore, if the target square is immediately above the red line, the score is

${\displaystyle S_{1}=(\varphi +2(\varphi ^{2}+\varphi ^{3}+\varphi ^{4}\ldots ))(1+\varphi +\varphi ^{2}+\varphi ^{3}+\ldots )}$.

At this point, observe another interesting property of ${\displaystyle \varphi }$, namely that ${\displaystyle \sum _{n=2}^{\infty }\varphi ^{n}=1}$. Applying this identity produces

${\displaystyle S_{1}=(\varphi +2)(1+\varphi +1)=(\varphi +2)^{2}=5+3\varphi \approx 6.85410\ldots }$.

If the target square is in the second row above the red line, every soldier is one space further from the target square, and so the score is

${\displaystyle S_{2}=\varphi S_{1}=3+2\varphi }$.

Similarly:

${\displaystyle S_{3}=\varphi S_{2}=2+\varphi }$,

${\displaystyle S_{4}=\varphi S_{3}=1+\varphi }$,

${\displaystyle S_{5}=\varphi S_{4}=1}$.

When a soldier reaches the target square after some finite number of moves, the ending configuration has score ${\displaystyle E=\varphi ^{0}+\epsilon }$, where ${\displaystyle \varphi ^{0}}$ represents the contribution of the soldier on the target square and ${\displaystyle \epsilon }$ represents the (small, but positive) contributions of the infinite number of soldiers that remain elsewhere on the plane.

Thus, we have shown that when the target square is in the fifth row above the infinite half-plane of soldiers, the starting configuration's score is exactly ${\displaystyle S_{5}=1}$; the ending configuration's score is ${\displaystyle E=1+\epsilon }$; and since no kind of jump ever increases the score, we must have ${\displaystyle S_{5}\geq E}$. This is a contradiction; Q.E.D., it is impossible for any soldier to reach a square in the fifth row after a finite number of jumps.