# Thread: OMG!! I gotta post this equation i have asking myself!

1. Ok, I'll explain this one, I have found a very curious equation and i'm gonna show it to you right now (it's not a very complicated one thought:

Ok, there are two simple, very obvious answers, which are 2 and 0, but i wondered for weeks if there were other solutions for this equation(sorry, currently i don't have the mathematic "tools" to answer that question)

2.

3. When you provide two unique variables in an algebraic expression, this implies two unique numbers. If we use the numbers you previously state, we find that and don't satisfy the equality , as and . So, the equality should be written as , which simplifies to .

The reason this works is because of a series of patterns that occur within number theory. One for example, which I discovered a few years ago, is such (I'm unaware whether or not this series has been discovered previously, and can only conjecture as far as my knowledge):

a series listing the ratios between the second power of any number and twice its value, or algebraically as such , starting at one will have a value of , and will increase by that value per term. Nowhere in this series is there a quotient of zero.

So for example, the following are the first few terms of the aforementioned series:

As you can see, at the second number of the series, the quotient is equal to one, which is the value derived at the number two. Thus, there's no deviation between squaring the number two and adding it twice. The reason zero works is that at the term of zero (the series I posted starts at one), the quotient is undefined, as its equation is , and so all such arguments are superflous.

4. I tried to generalize it and did some checks on the equation:

with

It seems to have always solutions and two of these are:

, , (and also when n=odd)

The rest of the roots are always imaginary.

So for there appear to be no more solutions than 0 and .

I guess DrRocket can cough up a proof for this within seconds. 8)

5. -2

6. Originally Posted by Incoming Dessert
-2

-2 doesn't work because one result is gonna be positive and the other is gonna be negative

7. Originally Posted by Ellatha
When you provide two unique variables in an algebraic expression, this implies two unique numbers. If we use the numbers you previously state, we find that and don't satisfy the equality , as and . So, the equality should be written as , which simplifies to .
Yeh, that's what i meant. I meant that X and Y were the same number, so if these two variables worth 0, it works, and when they worth 2 they work too.

8. Originally Posted by MathManiac
Ok, I'll explain this one, I have found a very curious equation and i'm gonna show it to you right now (it's not a very complicated one thought:

Ok, there are two simple, very obvious answers, which are 2 and 0, but i wondered for weeks if there were other solutions for this equation(sorry, currently i don't have the mathematic "tools" to answer that question)
Think of it as two surfaces in 3-space

z=xy and z-x+y. A solution is any point on the intersection of those two surfaces. There will be many such points.

9. If x and y are the same number, then it just simplifies to x*(x - 2) = 0, which only has the two solutions.

If they can be different, you can get it down to, IIRC, y = 1/(1 - 1/x), which has many (infinitely so) solutions.

10. And Nice work--thank you for sharing- for me this makes perfect sense though.
This is exciting news.
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11. Originally Posted by accountabled
I tried to generalize it and did some checks on the equation:

with

It seems to have always solutions and two of these are:

, , (and also when n=odd)

The rest of the roots are always imaginary.

So for there appear to be no more solutions than 0 and .

I guess DrRocket can cough up a proof for this within seconds. 8)
(My first post. Please excuse the notation struggles.)
I quite like this generalization by accountabled of the original question, which is about at the level of math I can handle these days, and just wanted to amplify on the subject of the complex roots. Using the complex variable z instead of x, and after allowing for the trivial solution z=0, the equation zn = nz reduces to z(n-1)=n, so that the solutions are the n-1 complex "n-1"-th roots of n. Thinking in terms of the polar, radius/angle representation for complex numbers z=reia , one can visualize that in the complex plane these are evenly spaced around the circumference of a circle with radius n1/(n-1) , with the first being the real root n1/(n-1) itself. Thus, as accountabled noted, if n is odd, so that n-1 is even, one of these n-1 roots will be the negative real -n1/(n-1) ; and furthermore if n-1 is divisible by 4 there will also be the pure imaginary roots in1/(n-1) and -in1/(n-1) . The simplest case of this is n=5, which yields the 5 solutions 0,51/4,-51/4,i51/4,and -i51/4 .

12. Please be careful with the search function. This thread is 2 years old.

Also, you might want to read the Tex sticky since it makes entering such notation much easier.

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