Recent content by S.Iyengar

Perfect !, I am happy that you have entered the scene sir. Really glad to see you again.

Did you try anything better ?

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 ?.

Yes, but given a surd, we can't bring that expression into that form. I hope if you see my latest edit you will understand what I mean

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...

No problem sir.. I am happy that you responded in a nice manner.

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.

Thank you again sir

Ok thank you for your infinite patience 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...

Thank you sir.. I think he multiplied both sides with p^{-a+1} . Thanks a lot

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...

What about the situation when a=k ?

This time the proof looks rigorous sir. Thank you.