Prove that for any non-zero positive integers [itex]b,s[/itex](adsbygoogle = window.adsbygoogle || []).push({});

[itex]a = \large \log_x \bigg(\dfrac{ ( -3 + x^{2b} ) \pm \sqrt{(3-x^{2b})^2-4x^{2b}(1-s^2)}}{2x^{2b}}\bigg)\notin \mathbb{Z}.[/itex]

The above expression comes from the result[itex]x^a = \dfrac{ ( -3 + x^{2b} ) \pm \sqrt{(3-x^{2b})^2-4x^{2b}(1-s^2)}}{2x^{2b}}.[/itex]

Edit : The [itex]x^a[/itex] present there comes from the root of quadratic equation in terms of [itex]x^a[/itex] which is

[itex]x^{2b}.x^{2a} +(3-x^{2b}) x^{a} + (1-s^2)=0 .[/itex]

There [itex]a \ge 1, b \ge 1, s \equiv 0 \mod 2, x[/itex] is a prime [itex]\gt3[/itex].

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

Thank you.

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# Prove that [itex]a[/itex] is not an integer

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