This link does not work anymore, can give alternative source?

Thanks, can you refer me to related paper/book/theorem (keywords/ author name to search) , specially for the first part? Also, if you can improve the headline of the post for future users, please do that.

@GerryMyerson that is a joke right?? I don't see any other reason for informing the trivial case $y^2=x^3$

@JoeSilverman thanks, those are from Pell's equations family, right? is there any other such Power Diophantine Equation family? BTW, is last line of your comment is correct? would you please check it again?

Can you explain why $\alpha^r$ is close to $\gamma^s$ (on page 13), at the end?

I have updated and change the question, can you suggest any academic to whom I can mail for solution?

not much help and disagree hugely about your comment about right or wrong... anyway bounty awarded to you for your time..

@YemonChoi any idea how to crack it? we are beginners, so plz help if possible.

thanks for yor time but at least provide rigorous proof of, $\implies \frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{(1-\eta)rd}}{\alpha^{1+rd}}$**(how it is possible ?$\eta<1/2$ makes things spooky!)** and $\alpha^r\approx \gamma^s$ (how?),that is what we are looking to know + learn (forget the application of corollary 1). If possible do that before 9 hours.

@GerhardPaseman dude! $\frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{rd}}{\alpha^{1+rd}}$ $\implies \frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{(1-\eta)rd}}{\alpha^{1+rd}}$**(how?)** is not a simple "algebraic consequence", the member who provided it deleted his answer because he could not prove it (he programmed and found positive result but could not prove it)... we both tried , probably there is a 'general theorem' involved that goes without mentioning in the number math..

@GerhardPaseman there is another "how" :) .... go for the detail answer and take 100 points!!!