What is the appeal of the Denon 103 cartridges?

Tubehead, The formula in the form you quoted is the same as mine, except you start with the end of my calculation.  It's just algebra.  But 99% of the time, one is going to use the formula starting with a known turns ratio in order to determine the reflected impedance across the transformer.  If you start with a known turns ratio, you square it (obviously), e.g., a 1:10 turns ratio will reflect an impedance which is equal to the resistance across the secondaries divided by 100 (10^2).  So, in that use of the formula, there is no need to take the square root of any number, albeit the square root is available rapidly with any decent calculator.  I agree that we really have no argument here.
Now, can you tell me why the "rule" for loading is 2X for a SUT and 10X for the pure resistive load of a high gain phono stage?  What's also odd to me: I have only ever seen this reasoning used among those who use the Denon DL103 series of cartridges.

Tubehead, you may have misspoke by accident. The formula relates the reflected impedance as inversely proportionate to the square of the turns ratio. (For a 47K resistor load, a 1:10 SUT will reflect 470 ohms to the cartridge, as you correctly stated. That’s 47K divided by 100.) There’s no need to worry about square roots. It’s been many years since I listened to a DL 103, but I don’t know why such a set up wouldn’t work just fine without any further messing around with the load resistance. Also, I have always wondered, and still cannot understand, why the goal for the load resistance should be any different when you are using an SUT, compared to when you are using a high gain phono stage. Perhaps someone else can enlighten me.