\((d,e,xy)\) is another solution to \(X^4-Y^4=Z^2\)
Proof:
Remember we started with \(A^4+B^4=C^4\) where \((a,b,c)\) was a solution. Using this, we created a new formula: \((a^2)^2=c^4-b^4\).
We let \(z=a^2\), \(x=c\), and \(y=b\). Through substitution, we get \(Z^2=X^4-Y^4\) where \((x,y,z)\) was a solution. This can be shown by substituting equivalent values in for \(x\), \(y\), and \(z\), and working backwards to get our original equation \(A^4+B^4=C^4\) with solution \((a,b,c)\).
The current formula is \(Z^2=X^4-Y^4\) with solution \((x,y,z)\). Through manipulation, we found \((y^2,z,x^2)\) is a primitive Pythagorean triple. Based on those properties and more algebra, we found \(x^2y^2=d^4-e^4\). So why is this a solution? Using the laws of exponents we can rewrite this as \((xy)^2 = d^4-e^4\).
Let \(xy=Z\), \(d=X\), and \(e=Y\). Through substitution we see our original formula \(X^4-Y^4=Z^2\) arise. Hence, \((d,e,xy)\) is another solution to\(Z^2=X^4-Y^4\).
Q.E.D.