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$2^{^{11}} a + 2^{^{11}} b + ab = 1$

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?

my_username
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    Hint: Consider the product $(a+2^{11})(b+2^{11})$ – jjagmath Sep 11 '21 at 15:33
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    Given $\space 2048(a + b )+ ab - 1 = 0\space $ At least one of $a,b$ is negative as Wolfram Alpha shows in the $16$ solutions here. Click more solutions to see your solutions in the 2nd set of $8$. – poetasis Sep 11 '21 at 23:52

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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)

Jorge
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