User:Bernard/Yin Needs Yang

Background
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Solution
Consider the equation

$$(N + 1)^2 = N^2 + (2N + 1)$$

Now suppose you had a sign with the numbers N^2, N + 1, and 2N + 1.

Then you could make things work out using either

$$\sqrt{N^2} = (2N + 1) - (N + 1)$$

or

$$\sqrt{(N^2 + (2N + 1))} = (N+1)$$

In the example pictured below, we have the numbers for N = 10, namely 100, 11, and 21. Therefore the sign yields two "solutions",

$$\sqrt{100} = 21 - 11$$

and

$$\sqrt{100 + 21} = 11$$

What I think is really cool here is that a kind of mirror image sign results if N is negative. Sounds crazy, but stick with me. Suppose $$N = -10$$. Then $$N^2 = 100$$ and $$(2N - 1) = -19$$ and $$(N - 1) = -9$$. So with the numbers 100, 19 and 9 we would have

$$\sqrt{100} = -9 - (-19)$$

...or, said another way, 19-9

and $$\sqrt{100 - 19} = -(-9)$$

...or, said another way, 9

So that means the sign we have in the pictures down below has a sister sign from the dark (negative) side, 100, 19, and 9.

One proposal is that there be special Prize* for signs from the dark side that match signs we already from the light side. Or, for that matter, a light side sign that matches one from the dark side. In this case, whoever finds a sign with 100, 19, and 9 would be recognized for finding the dark match to the sign in this email. Another option would be to create a sign type for these, with the suggested name of Doppelgänger.

And there's more - while the quest for 100, 19, and 9 goes on, somebody else might find, say, the dark side from N = 8. ( That would be 64, 7, and 15.) That would start another frantic search for the bright side of N = 8. (That would be 64, 9, and 17.)

Suggestion

 * I suggest we call it the Thinglestad Prize, which might be a special logo on the picture of the sign. After a few years, we might be able to have some kind of virtual dinner to honor the Thinglestad Prize awardees?