Considering the Fermat's little theorem, considering divisibility on both sides by the nos that occur on LKS and RHS, we get that $x \equiv 7 ( \bmod 12)$ and $y$ is even..... First few solutions $(x,y)$ = $(1,0)$ and $(7,4)$..... What further????

Can somebody please post a proper solution fast!!! I want to know whether my steps and answer( till now) is correct! Help needed urgently!!

Rewrite the equation as $6 \cdot 3^z +1 = 7 \cdot 5^y$, where $z=x-1$. Aside from $(0,1)$, we quickly get by examining modulo 3 and 5 that $x,y$ are even. Thus we are looking at solutions of $7A^2-6B^2=1$. These solutions must be in the form $(a,b)$, where $a\sqrt7+b\sqrt6= (\sqrt7+\sqrt6)^n$, and $n$ is odd. The even terms of the binomial expansion of $(\sqrt7+\sqrt6)^n$ are integral multiples of $\sqrt6$, and all but the first of these (without the $\sqrt6$ part) are necessarily divisible by 3. So $n$ must be divisible by 3 as well. Looking at $n=3$ yields $(25,27)$, clearly a solution. If $n>3$, then $a\sqrt7+b\sqrt6= (25\sqrt7+27\sqrt6)^m$, for integral $m$. Looking at the even terms (without the $\sqrt6$ part) modulo 81 yields that $m$ is divisible by 3 as well. Continuing this way, we must have $a\sqrt7+b\sqrt6= (25\sqrt7+27\sqrt6)^{3^r}$. Again, looking at the even terms (without the $\sqrt6$ part), we see that all but the first are divisible by $3^{r+4}$, a contradiction. So the only solutions of the original equation are $(1,0),(7,4)$.

Is there another solution ?

A special case of China 2005 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=155821&sid=2d6b0b213258afcc048c0074ee6bd0b1#p155821

utkarshgupta wrote: A special case of China 2005 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=155821&sid=2d6b0b213258afcc048c0074ee6bd0b1#p155821 No, they are obviously different.

utkarshgupta wrote: Is there another solution ? Yes . $2*3^x+1=7*5^y$ Set $x=2z+1$ , $y=2w$ (it is easy to prove it). Transform the equation as.. $(3^{2z+1})^2+7*(5^{w})^2=(3^{2z+1}+1)^2$ The solutions of equation $x^2+py^2=z^2$ for $y$ odd and prime $p>2$ are $x=\dfrac{p-1}{2}*(s^2+t^2)+(p+1)st$ $y=s^2-t^2$ $z=\dfrac{p+1}{2}*(s^2+t^2)+(p-1)st$ It is easy to find the acceptable pairs of $(x,y)$ . Δημήτρης