I'm going to take a guess as to what OmnipotentEntity sees as the "easy method". I'm 90% sure it is this:
![](https://i.imgur.com/HiJtP5y.png)
Let y and z be as shown in the image above based on the subdivisions of the bottom edge. Then:
(2x)
2+y
2=145 (A)
(5x)
2+(y+z)
2=689 (B)
(4x)
2+(z+3)
2=377 (C)
Now let's declare that x,y,z are natural numbers. (Just do this. Don't think about it.)
Equation (A) says that 145 = 29*5 is a sum of two squares. Checking shows that 145 = 8
2+9
2 = 12
2+1
2 (you can also use 29 = 3
2+2
2 and 5 = 2
2+1
2, and the
Brahmagupta-Fibonacci identity to get the above sums of two squares). The first gives x=4, y=9, and then in (C) z=8, and (B) checks out. The second gives x=6, but then (C) has no solution for z, so reject that. Therefore x=4.
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If you are unsatisfied with the above argument, specifically the declaration that x,y,z are natural numbers, note that no other argument I've found is interesting. You are free to solve it in real numbers by squaring equation(s) a few times to get rid of radicals.
WolframAlpha will happily tell you that the only solutions are x=±4.