# Find all positive integer solutions of $(a,b,n)$ s.t. $(2^a-1)(2^b-1)=n^2$

So recently I've received a problem (title) and I have no idea how to solve it. By easy observation, $$(a,b)=(m,m),(6,3),(3,6)$$ are the cases I've found, and no others.

So I try to let $$kA^2=2^a-1$$ and $$kB^2=2^b-1$$, where $$k = \gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1$$. Then, another equation appears:$$\dfrac{2^a-1}{2^{\gcd(a,b)}-1}=A^2$$ the other one is similar. This expression is a bit ugly so I let $$\gcd(a,b)=x$$ , $$a=xy$$. The equation then becomes: $$\dfrac{2^{xy}-1}{2^x-1}=A^2$$ which is much nicer. The solutions I have found for now (which is according to the (a,b) solutions above) are $$(x,y)=(x,1),(3,2)$$. Also, by modulus and some algebra, if there are more solutions we just found, then $$x\ge3$$ and $$y\ge 5$$. Well that's it. I can't go further more. Please give me some idea or even tips or solutions. Also, no computer is allowed. Thank you.

• Isaac, There is a known fact that says Mersenne numbers which have the form $2^n-1$ can not be any power more than 1(including 2) of an integer. So you have to find common factors in two terms, like what you found $(63, 7)=7$. Nov 17 '20 at 10:42
• Zsigmondy's theorem may be of interest (actually Bang's theorem, which is the special case with $a=2$ and $b=1$). Nov 17 '20 at 11:25
• @Servaes I see. So it directly shows that $xy=6$ or $y=1$ in this case? Nov 17 '20 at 12:13
• No; if $p$ divides $2^a-1$ but not $2^b-1$, then you might still get a solution if $p^2$ divides $2^a-1$. So this doesn't solve the problem yet, but does put strong restrictions on possible solutions. Nov 17 '20 at 12:19
• I'm wondering whether you can use the fact that an odd square $\equiv 1 \bmod 8$. Nov 18 '20 at 17:12