\(d\) is smaller than \(x\), which tells us a new solution with a smaller first value can be found from any given solution.
Proof:
Recall that \(x^2=d^2+e^2\). We know that \(e^2>0\). Thus is follows: \[e^2>0\] \[e^2+d^2>d^2\] \[x^2>d^2\] \[x^2-d^2>0\] Factoring the left hand side we see: \[(x-d)(x+d)>0\] By their definition, \(x\) and \(d\) are both greather than 0. Therefore their sum is also greater than 0, so we can safely divide by \(x+d\) on each side of the inequality. Doing so reveals: \[x-d>0\] \[x>d\] Therefore, \(d<x\). Having shown this, we have found a smaller integer from which a new solution can be made. Q.E.D.