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I found a curve, in which some function has at least two expressions, which differ infinitely much!! Is there any error in the thoughts?
The curve is defined by

"\begin{equation} Ax²+Bx+C=u²,\end{equation} \begin{equation} A'x²+B'x+C'=v²,\end{equation}
provided either $A$ and $A'$, or $C$and $C'$, are squares; it may be assumed that the left-hand sides have no common zero on the projective straight line, since otherwise this would define a curve of genus 0."
(Changed from the book Number theory, An approach through history from Hammurapi to Legendre by A.Weil)

Further define four points $P_{+,+}$,$P_{+,-}$,$P_{-,+}$,$P_{-,-}$, respectively by $(0,1,a,a')$,$(0,1,a,-a')$,$(0,1,-a,a')$,$(0,1,-a,-a')$. Now we consider the function $f$=$a'u+av-t_0$, with $t_0$ given by $(A'B-AB')/(2aa')$. By (I), we have the expression for $f$:

Since $(a'u+av)(a'u-av)=(A'B-AB')x+(A'C-AC')$, by dint of some easy calculations,

Now, substituting the values previously defined, we find that $f$ should actually go to infinity at $P_{+,+}$, and at $P_{-,-}$. Nevertheless, direct substitution tells us that $f$ does not go to infinity at these points!! How can this be true?
Ever since the times of the high school, the identities of different expressions for functions have always been a useful (if not the only) means of obtaining important results (to me). But I seem to fail to manipulate correctly here. One possible explanation for this phenomenon IMO is in the process of embedding the curve in the projective line, but I figured not out where exactly it went wrong.

Hope the formulation is not too confusing or obscure.

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Personally, I got bored reading your introduction, which doesn't contain any interesting information, before I even got to the question. In my opinion, you are more likely yo get people to read your question if you stick to mathematics (and use latex). – Alex B. Aug 16 '11 at 8:01
Like Alex B. said. – Did Aug 16 '11 at 8:04
Thank you for giving me the suggestions; I will try to emend. – awllower Aug 16 '11 at 10:55
There is no introduction now; if it is concerned with the latex, please teach a young boy who knows little about latex. Sorry for any inconvenience; I just want to know the answer, and thank you all. – awllower Sep 1 '11 at 13:18
up vote 1 down vote accepted

I finally found an answer to this question. The reason for the points to be poles is exactly as I said before, and as to the zeros, the reason is simple enough:Since the points are in the projective space, the constant terms are deleted, and hence they are the zeros.
Moreover, as to the reason to choose some specific points instead of others, it is just because here we are looking for some rational divisors, and so we must ensure that they are the linear combinations of prime rational divisors, which makes sure that they ought to be chosen this way!!
Indeed, the function not at all has as values which disagree with each other at any point on the curve given. So, I guess this really answers the question, posed by me before!!

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I guess I did receive some help in any case from revising it!! – awllower Oct 5 '11 at 13:12

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