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Does anyone know how to solve the following equation? $(DX)^2-Y^2=(\sqrt{2}-7)^5$ Is there any special name for these equations?

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Over what number system? Diaphontine equations usually are solved over the integers or naturals. With an irrational on the right, that won't happen. Why are you using $DX$? It could be just $X$, couldn't it? –  Ross Millikan Dec 7 '11 at 3:14
    
All algebraic numbers.I use DX because D and X have to be raised to $2$ –  Vassilis Parassidis Dec 7 '11 at 3:31
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Does anyone know how to get Vassili's accept rate up over 20%? –  Gerry Myerson Dec 7 '11 at 4:09
    
Gerry how a universal Diophantine Equation looks like an equation which includes all algebraic numbers? –  Vassilis Parassidis Dec 7 '11 at 4:19
    
That great authority en.wikipedia.org/wiki/Diophantine_equation says Diaphontine equations "allows the variables to be integers only" –  Ross Millikan Dec 7 '11 at 5:06

2 Answers 2

Let $D$ be any algebraic number you want. Let $X$ be any algebraic number you want. Let $Y$ be either of the two solutions to $Y^2=(DX)^2-(\sqrt2-7)^5$. That gives you all the solutions in algebraic numbers, just what you asked for.

The special name for this equation is "trivial".

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Gerry give me numbers for Y and DX only numbers accepted –  Vassilis Parassidis Dec 7 '11 at 4:24
    
OK: $D=X=0$, $Y=(7-\sqrt2)^{5/2}$. I've done my bit - now you go accept answers to some of your questions. –  Gerry Myerson Dec 7 '11 at 5:13
    
Gerry these equations are far more complicated than you think. In order to understand them you have to know the way in which are formulated. First you have three variables from them you make up a number of groups. Using addition, subtraction and multiplication of these groups you get a final result. This result you separate into DX and Y to obtain the general solution for different powers of the binomial. In my opinion none of you can present a solution for n=7 for this reason please have one of the best number theorist to look at these equations. –  Vassilis Parassidis Dec 8 '11 at 3:27
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You don't know what you are talking about. There isn't even an $n$ in your question to set equal to 7. Now play fair: you asked a question, and I answered it, clearly, correctly and completely. You asked another question, and I answered that one, too, clearly, correctly, and completely. Now go accept answers to some of your questions. –  Gerry Myerson Dec 8 '11 at 5:23

It is just a quadratic equation. $Y=\pm\sqrt{(DX)^2-(\sqrt{2}-7)^5}$

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You can't solve this equation without first knowing the values of D and X. –  Vassilis Parassidis Dec 8 '11 at 3:04
    
Nonsense. Ross' formula works no matter what $D$ and $X$ are (and so does my solution). –  Gerry Myerson Dec 8 '11 at 5:25
    
Let's have the following groups of the three variables: –  Vassilis Parassidis Dec 8 '11 at 5:45
    
Let's have the following groups of the three variables: $a=4y^2\sqrt{x}+4xy$ $b=y^2+\sqrt{x}y+x$ $c=2y^2+2y\sqrt{x}$ $f=xy$ $g=y^4+x^2$ $h=\sqrt{x}+y$. It is allowed to use addition, subtraction and multiplication of these groups and the final result has to be separated into two parts, one equal to $DX$ and the other equal to $Y$. From these we get the global solutions of this equation $(DX)^2-Y^2=(x-y^2)^5$. x and y can be any algebraic numbers. These are what I call universal Diophantine equations. –  Vassilis Parassidis Dec 8 '11 at 6:00
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I don't understand the motivation or interest of this definition. –  Ross Millikan Dec 8 '11 at 14:39

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