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Motivation :

The motivation is to show that the equation $x^{2b}.x^{2a} +(3-x^{2b}) x^{a} + (1-s^2)=0 $ has no solutions in integers for any values of $x,b,a,s$ ( choosen as per the constraints imposed below ) . So to prove that there are many ways. That needs a great depth of knowledge about diophantine equations. But I have tried an alternate method of finding a contradiction. That is to believe initially that the equation has solutions and later on meddle with the quadratic formula and obtain a contradiction.


We have this equation with us $$x^{2b}.x^{2a} +(3-x^{2b}) x^{a} + (1-s^2)=0 .$$There $a \ge 1, b \ge 1, s \equiv 0 \mod 2, x$ is a prime $\gt3$.

So we need to obtain the contradiction based upon the given constraints. There are few ways of obtaining the contradiction which I worked out. I am stuck in the middle. Please do post some ideas to let me go further. The statement to be proved is

Prove that there are no such integers $a,x,b,s$ such that $$x^{2b}.x^{2a} +(3-x^{2b}) x^{a} + (1-s^2)=0 .$$

Proof By Contradiction 1 :

We assume that the given statement has solutions. Then we proceed for the quadratic roots formula and there we need to obtain some contradiction. The contradiction we are going to get is that the assumed $a$ which is an integer, turns out to be a floating point number at-last. $$x^a = \dfrac{ ( -3 + x^{2b} ) \pm \sqrt{(3-x^{2b})^2-4x^{2b}(1-s^2)}}{2x^{2b}}$$ $$\big \Downarrow $$

$$a = \log_x \left(\dfrac{ ( -3 + x^{2b} ) \pm \sqrt{(3-x^{2b})^2-4x^{2b}(1-s^2)}}{2x^{2b}}\right)$$

Then one possible way of obtaining the contradiction is showing $a \notin \mathbb{Z}$.

I don't know whether $a$ turns to integer or not . Any counter-examples that make $a \in \mathbb{Z}$ are highly appreciated. I have substituted some random values and got the numerator always less than denominator, and there by creating an obstruction for $a$ to be an integer. So can we have some comparisons on the numerator and denominator ? .

Proof By Contradiction 2:

If the equation has a solution, then its discriminant should be a square ( square of $k$ ) . So taking its discriminant into consideration, we get

$$x^{4b}+x^{2b}(4s^2-10)+9=k^2.$$

So we need to obtain some contradiction from here. My attempts are

  • We have $x^{2b}(x^{2b}+4n^2-10)=k^2-9$ . So $x^{2b} | (k^2-9) \implies k=\mathfrak{C}x^{2b} \pm 3$ for some positive $\mathfrak{C}\in \mathbb{Z}$ , from there by using some modular arithmetic techniques we need to prove some thing.
  • Another way of doing it is fitting the $ x^{4b}+x^{2b}(4s^2-10)+9$ between two squares, such that $$ (m-1)^2 \lt x^{4b}+x^{2b}(4s^2-10)+9 \lt (m+1)^2 $$ then comparing with $k^2$, I know how to prove the contradiction.

Please do let me know how to proceed further.

Thanks a lot for everyone who read this post with patience. I would be happy if someone comes with a new contradiction, or the one's mentioned above.

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    I'm a little confused -- here $x^a$ is *defined to be* the set in that expression, rather than $x$ raised to the $a$ power or $\exp{(a\ln{x})}$?2012-07-09
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    What is $x$? Is it any real solution to the equation you wrote down? Where does your problem com from (you didn't make this question up like that I suppose); the quadratic formula seems to be involved, but it's not obvious how?2012-07-09
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    @MarcvanLeeuwen : Its evident that you can manipulate and find out you get a quadratic equation, anyway I have edited it, thank you.2012-07-09
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    @Neal : I hope now my question is clear after the edit.2012-07-09
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    There are some counter-examples out there that make $x^a \in \mathbb{Z}$ (which is what you asked for), but those I've found make $a \notin \mathbb{Z}$. Two examples: $b=2, s=328, x=4$ and $b=3, s=36, x=2$.2012-07-09
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    @PeterPhipps : My dear Peter, I thank you for the efforts you made to produce a counter-example. But I have clearly stated $x$ to be a " PRIME " greater than $3$. May be my write-up is not clear so that you have missed reading it.2012-07-10
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    No, the problem is mine for misreading it.2012-07-10
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    The problem as currently stated makes no sense. It starts, "for any positive integer $a$..." and then asks to prove that $a$ is not an integer. Please take a little more care in stating the question.2012-07-11
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    @GerryMyerson : Dear sir, Thanks a lot for informing me. I was looking for a contradiction that will arise making $a \notin \mathbb{Z}$ . Now I re-edited the question adding "Contradiction Required" . We start with $a$ as an integer and later on we get a contradiction that $a$ is not an integer if that equation holds good. Do you have any other suggestions ?2012-07-11
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    Huh? If $a$ is an integer (as postulated in the 1st statement in the question) then of course there is a contradiction in anything that ends with $a$ not being an integer, as it says at the end of your last arrow.2012-07-11
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    You can't prove that $a$ is not an integer if you start, as you do, with the assumption that $a$ is an integer --- unless, of course, you manage to prove that mathematics is inconsistent.2012-07-12
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    @GerryMyerson : I made a lot of improvements, now it may look nice.2012-07-16
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    It makes sense - but it's utterly unmotivated. Why would anyone be interested in such an equation? How could it possibly arise? How did you come upon it?2012-07-16
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    @GerryMyerson : There is no motivation sir. My plan was to prove that the equation has no solutions.2012-07-27
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    @GerryMyerson : I have now added the motivation sir. I beg you to use '@' while posting comments, I didn't see the comment from many weeks, and that's why I didn't edit it so far. Now I have done by your kind suggestion. Thank you for that.2012-07-27
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    @Iyengar I think it still lacks some motivation, in the sense of "Why this particular equation?". I can't see where it arises from, or why one wants to investigate such an unusual-looking equation.2012-07-27
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    @OldJohn : Its simply a Diophantine equation. I was asked to solve it by a professor. I don't think there will be some motivation behind choosing some equations. For example, why does we need to know $x^n+y^n \neq z^n $ when $n\gt2$ ? .2012-07-27
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    Since you posted the question, you should see all comments even if no one uses the at-sign. You can just go to your user page and click on "responses". And if you were asked to solve it by a professor, then I assume the professor wants you to solve it, and doesn't want me to solve it. Otherwise, the professor would have posted it here, instead of giving it to you. You might ask the professor the motivation behind the question.2012-07-27
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    @GerryMyerson : Yes, thank you for teaching me the responses trick, which I don't know previously. Its true that professor wanted me to solve but I was stuck some-where. I can't proceed further. I never asked you to solve it completely , but instead I wanted someone to post some ideas/hints in that direction.2012-07-27

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