XOR-Equation C
We use $x\oplus y$ for the bitwise XOR of $x$ and $y$.
Define the XOR-product of $x$ and $y$, denoted by $x \otimes y$, similar to a long multiplication in base $2$, except that the intermediate results are XORed instead of the usual integer addition.
For example, $7 \otimes 3 = 9$, or in base $2$, $111_2 \otimes 11_2 = 1001_2$: $$\begin{align*} \phantom{\otimes 111} 111_2 \\ \otimes \phantom{1111} 11_2 \\ \hline \phantom{\otimes 111} 111_2 \\ \oplus \phantom{11} 111_2 \phantom{9} \\ \hline \phantom{\otimes 11} 1001_2 \\ \end{align*}$$ We consider the equation: $$\begin{align} (a \otimes a) \oplus (2 \otimes a \otimes b) \oplus (b \otimes b) = c \otimes c \end{align}$$
Define the XOR-product of $x$ and $y$, denoted by $x \otimes y$, similar to a long multiplication in base $2$, except that the intermediate results are XORed instead of the usual integer addition.
For example, $7 \otimes 3 = 9$, or in base $2$, $111_2 \otimes 11_2 = 1001_2$:
For example, $(a, b, c) = (1, 2, 1)$ is a solution to this equation, and so is $(1, 8, 13)$.
Let $F(N)$ be the number of solutions to this equation satisfying $0 \le a \le b \le N$. You are given $F(10)=21$.
Find $F(10^7)$.