Dear sir,
But that may not seem to work perfectly. If the term also gives some decimals and make it cancel both the sides then it would become an integer.
Yes may be you are right .
I did something else to make it appear as follows . Log (a/b) = log a - log b. It makes a term Log_x(2) in the expression ( that arises when we expand \log_x(2x^{2b}). So that Log_x(2) may contribute to some decimal part . How about this ?.
Prove that for any non-zero positive integers b,s
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}.
The above expression comes from the result x^a = \dfrac{ ( -3 + x^{2b} ) \pm...
Another small misunderstanding sir. You have written that , k^2p^a+2k=2kp^{a+n}+2p^n-p^{2n}+3 and have said that, all the other terms are divisible by p^n. But we can't write that, given there is a term k^2p^a on the L.H.S
Thanks a lot again sir.
But I find another flaw sir.
You wrote that
r^2-1 = p^a(2rp^n-p^{2n}+3) = (r+1)(r-1) [2]
So if r > 1, r = kp^a \pm 1, some k > 0 [3]
So the fact that [2] \implies [3] is not correct. Because it follows from the argument that either (r-1) \rm{or} \ (r+1) | (2rp^n-p^{2n}+3) .
So the...
Dear sir,
Here you considered that
3p^{a+2n} \le 6p^{a+n} + 9p^a + 4p^{2n} \\
3p^{1+2n} \le 6p^{1+n} + 9p + 4p^{2n-a+1} \\
3p^{1+2n} \le 6p^{1+n} + 9p + 4p^{2n}\\
(3p-4)p^{2n} \le 6p^{n+1} + 9p
There in the second step you have substituted a=1 and then proceeded further. So what...