---

Problem 945

---

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 $2$, $111_2 \otimes 11_2 = 1001_2$:
$$
\begin{aligned}
\phantom{\otimes 111} 111_2 \\
\otimes \phantom{1111} 11_2 \\
\hline
\phantom{\otimes 111} 111_2 \\
\oplus \phantom{11} 111_2 \phantom{9} \\
\hline
\phantom{\otimes 11} 1001_2 \\
\end{aligned}
$$
We consider the equation:
$$\begin{aligned}
(a \otimes a) \oplus (2 \otimes a \otimes b) \oplus (b \otimes b) = c \otimes c
\end{aligned}$$
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 $F(10^7)$.

---

异或方程（三）

用$x\oplus y$表示$x$和$y$按位异或的结果。

考虑$x$和$y$的$2$进制长乘法，但将中间结果的相加替换为按位异或，定义其结果为$x$和$y$的异或积，并记作$x \otimes y$。

例如，$7 \otimes 3 = 9$，或在$2$进制下，$111_2 \otimes 11_2 = 1001_2$：
$$
\begin{aligned}
\phantom{\otimes 111} 111_2 \\
\otimes \phantom{1111} 11_2 \\
\hline
\phantom{\otimes 111} 111_2 \\
\oplus \phantom{11} 111_2 \phantom{9} \\
\hline
\phantom{\otimes 11} 1001_2 \\
\end{aligned}
$$
考虑方程：
$$\begin{aligned}
(a \otimes a) \oplus (2 \otimes a \otimes b) \oplus (b \otimes b) = c \otimes c
\end{aligned}$$
例如，$(a, b, c) = (1, 2, 1)$是该方程的一个解，$(1, 8, 13)$则是另外一个解。

记$F(N)$为满足$0 \le a \le b \le N$的该方程解的数量。已知$F(10)=21$。

求$F(10^7)$。

---