Integer solutions to $2^{^{11}} a + 2^{^{11}} b + ab = 1$ $2^{^{11}} a + 2^{^{11}} b + ab = 1$
By guessing that $a+b = 0$, I was able to find the solutions (a, b) = (-63, 65), (65, -63). Is there any practical way of finding other solutions to the equation?

Your equation is equivalent to the following
\begin{align*}
2^{22} + 2^{11}a + 2^{11}b +ab = 2^{22}+1 \\
(2^{11} + a)(2^{11} + b) = 2^{22} +1
\end{align*}
Now lets factor $2^{22} +1$ using Sophie Germain identitity
\begin{align*}
2^{22} + 1 =4(2^5)^4 +1^4 = (1 + 2^{11} - 2^6)(1 + 2^{11} + 2^6) \\
2^{22} + 1 = 1985 \cdot 2113 = 5 \cdot 397 \cdot2113
\end{align*}
These numbers ($5, 397, 2113$)are prime factors, so they divide either $2^{11} + a$ or $2^{11} + b$, but not both of them. There are very few posibilities (be careful and consider that both factors can be negative)
