A Prime Number of Rails

1. A right-angled triangular field has one side exactly $47$ rails long, and all three sides are a whole number of rails. How many rails are needed to fence it right round?

2. What if that side were $48$ rails instead?

Solution

Part 1. We need a right triangle with whole-number sides, one of them $47$. As $47$ is far too short to be the hypotenuse, take it as a leg, with other leg $b$ and hypotenuse $c$. Then $$47^2 = c^2 - b^2 = (c - b)(c + b).$$ Since $47$ is prime, $47^2 = 2209$ can only be written as $1 \times 2209$, so $c - b = 1$ and $c + b = 2209$, giving $c = 1105$ and $b = 1104$. The sides are $47$, $1104$, $1105$ (and indeed $47^2 + 1104^2 = 1105^2$), so the fence needs $$47 + 1104 + 1105 = 2256 \text{ rails.}$$ A prime leg pins the triangle down completely: there is only one such field.

Part 2. With $48$ the trick fails, because $48^2 = 2304$ splits in many ways into two factors of the same parity, and each gives a different triangle, for example $$\begin{gathered} (36, 48, 60), \quad (20, 48, 52), \quad (48, 55, 73),\\ (48, 64, 80), \quad (48, 90, 102), \quad \dots, \quad (48, 575, 577). \end{gathered}$$ So the question now has no single answer; the field could be fenced in many different ways. A prime side fixes a right triangle, a composite side does not.