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I had a question in my mind from many years. we generally present the trivial solutions to Diophantine equations. Diophantine equations usually always have some sort of trivial solution (if you allow zero even Fermat’s equation has trivial solutions: $0^n+x^n=x^n$). I suppose the hard question is whether there are non-trivial solutions? The question of non-trivial solutions is already tough for polynomial Diophantine equations and requires detailed knowledge in algebraic and arithmetic geometry, modular forms, etc (see Wiles’ proof of Fermat’s last theorem).

What I am looking for, can we find non-trivial solutions to any polynomial Diophantine equations? If yes, what are the method so far existing?

Thanks in advance to all members of mse.

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Its modular forms, please fix it. @gandhi – Iyengar Apr 3 '12 at 9:43
@Iyengar!I think now the edited one is much clear – gandhi Apr 3 '12 at 10:30
There was a fairly similar (although more focused) question on this site before:… – Alex B. Apr 3 '12 at 15:26
@gandhi: The solution to Hilbert's $10$th problem still leaves many interesting questions. Is there an algorithm for two-variable equations? Probably. For three-variable equations? For cubics in many variables? – André Nicolas Apr 3 '12 at 15:27

This question is Hilbert's 10th problem, from Hilbert's famous list of 23 problems promulgated 112 years ago in 1900.

The work of Julia Robinson, Martin Davis, Hillary Putnam, and Yuri Matijasevich, culminating in 1970, showed that it lacks an algorithmic solution.

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Sorry to say this, I think that there is some difference between diophantine equations and polynomial diophantine equations which OP is referring to . – Iyengar Apr 3 '12 at 12:55
@Iyengar : You'll need to explain. What Hilbert's 10th problem deals with is polynomials $p(x_1,\ldots,x_n)$ in several variables with integer coefficients. The problem is to find integer solutions. These are "polynomial Diophantine equations" since $p$ is a polynomial. Your answer seems to assume that polynomial solutions are sought. But I see nothing int he posting clearly indicating that. – Michael Hardy Apr 3 '12 at 15:16
See the comments above . – Iyengar Apr 4 '12 at 11:26

A polynomial Diophantine equation is an indeterminate polynomial equation whose solutions are restricted to be polynomials in the indeterminate. A Diophantine equation, in general, is one where the solutions are restricted to some algebraic system, typically integers

So you can refer to this wiki article and also this one for a background. Apart from that there is another article that serves a good start. For those who can access IEEE journals, this one is a very nice one.

Thank you and all the best.

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! I am not happy with your solution as it is irrelevant to my question in many aspects. Please...excuse and leave this. May be some other member will answer. Anyhow, Thank you so much for attempt. – gandhi Apr 3 '12 at 15:24
I see: There is a Wikipedia article that uses that terminology. That is different from what I would have thought the term meant. – Michael Hardy Apr 3 '12 at 18:29
@gandhi : I never bother about your happiness ( its not my duty and I never care too ) , but I only concentrate whether I am answering in a perfect manner or not, Hilbert's 10nth problem considers Diophantine equations, which are different from Polynomial Diophantine equations, if you think that the both were the same, why would there be a need to introduce two terms then . Let me give the precise conclusion ... CONTD. – Iyengar Apr 4 '12 at 11:18
CONTD.... A Diophantine equation is an indeterminate polynomial equation that allows the variables to be integers only, where as coming to the polynomial Diophantine equation it is an indeterminate polynomial equation whose solutions are restricted to be polynomials in the indeterminate. So in the case of polynomial diophantine equations, the solutions are polynomials themselves, see this , Now did you realize the difference between both of them ? @MichaelHardy – Iyengar Apr 4 '12 at 11:21
@Iyengar : Defining "polynomial Diophantine equation" in the way you have is just one convention. You ask: "if you think that the both were the same, why would there be a need to introduce two terms then[?]". There is a simple reason: One might want integer solutions to something like $x^y - y^x = 2$, and then one calls it a "Diophantine equation" because one seeks only integer solutions. But it's not a "polynomial Diophantine equation" since the expression whose zeros are sought is not a polynomial. – Michael Hardy Apr 4 '12 at 17:29

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