If you know the Diagonal and Area of a Rectangle, can you find the sides of the rectangle? If you know the Diagonal and Area of a Rectangle, can you find the sides of the rectangle? 
I was doing strange math yesterday, and I can across the realization that two different rectangles can’t have the same area and same diagonal. But I haven’t been able to solve a equation that shows this. I also haven’t be able to prove myself wrong? So maybe higher math gods will help?
$$\text{Diagonal}=\sqrt{x^2+y^2}$$
$$\text{Area}=xy$$

Sure:
You have $xy = A$ a known constant.
And $\sqrt{x^2 +y^2} = d$ a known constant.
So just substitute.
$x = \frac Ay$ (assuming $y\ne 0$ which if the area is positive must be so... or we could do $y = \frac Ax$.... it doesn't matter.
And $\sqrt{(\frac Ay)^2 + y^2} = d$.
So $(\frac Ay)^2 + y^2 = d^2$
So $A^2 + y^4 =d^2 y^2$
$y^4 - d^2y^2 +A^2 =0$.  Use the quadratic formula:
$y^2 = \frac {d^2 \pm \sqrt{d^4-4A^2}}2$  We know $y$ is positive
So $y =\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}$ and $x =\frac {A}{\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}}=\frac {A}{\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}}\frac {\sqrt{\frac {d^2 \mp \sqrt{d^4-4^2A}}2}}{\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}}=\frac {A}{\sqrt{\frac {d^4-(d^4-4A^2)}4}}\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}=\frac AA\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}=\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}$
....
So if we take the rectangle with sides $3$ and $4$ and area $A=12$ and diagonal $d=5$ and pretended we didn't know the sides we'd have.
$y = \sqrt{\frac {5^2+\sqrt{5^4-4*12^2}}2}=\sqrt{\frac {25+\sqrt{(25+2*12)(25-2*12)}}2}=\sqrt{\frac{25+\sqrt{49*1}}2}=\sqrt{\frac {25+7}2}=\sqrt{16}=4$
And $x = \sqrt{\frac {5^2-\sqrt{5^4-4*12^2}}2}=\sqrt{\frac {25-\sqrt{(25+2*12)(25-2*12)}}2}=\sqrt{\frac{25-\sqrt{49*1}}2}=\sqrt{\frac {25-7}2}=\sqrt{9}=3$
