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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


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 $\exp{(a\ln{x})}$? –  Neal Jul 9 '12 at 12:17
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? –  Marc van Leeuwen Jul 9 '12 at 12:23
No, the problem is mine for misreading it. –  Peter Phipps Jul 10 '12 at 9:20
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. –  Gerry Myerson Jul 11 '12 at 9:34
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. –  Gerry Myerson Jul 12 '12 at 0:17

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