10. Let $f(x)$ be a function defined on the set of integers, satisfying the conditions
(1) $f(1)=1, f(2)=0$;
(2) For any $x, y$,
$$
f(x+y)=f(x) f(1-y)+f(1-x) f(y) \text {. }
$$

Then $f(2015)=$ . $\qquad$
10. $\pm 1$.

In condition (2), let $x=0$, then
$$
f(y)=f(0) f(1-y)+f(1) f(y) .
$$

By $f(1)=1$, we know
$$
f(0) f(1-y)=0 \text {. }
$$

In the above equation, let $y=0$, then
$$
f(0) f(1)=0 \Rightarrow f(0)=0 \text {. }
$$

In condition (2), let $x=1,-1,2$ respectively, we get
$$
\begin{array}{l}
f(y+1)=f(1) f(1-y)+f(0) f(y) \\
=f(1-y), \\
f(y-1)=f(-1) f(1-y)+f(2) f(y) \\
=f(-1) f(1-y)=f(-1) f(y+1), \\
f(y+2)=f(2) f(1-y)+f(-1) f(y) \\
=f(-1) f(y) .
\end{array}
$$
$$
\text { By } f(y-1)=f(-1) f(y+1)
$$
$$
\begin{array}{l}
\Rightarrow f(y)=f(-1) f(y+2) \\
\Rightarrow f(y)=f^{2}(-1) f(y) \\
\Rightarrow f(-1)= \pm 1 .
\end{array}
$$

If $f(-1)=1$, then $f(y+2)=f(y)$.
By condition (1), we know
$$
f(x)=\left\{\begin{array}{l}
1, x \text { is odd; } \\
0, x \text { is even. }
\end{array}\right.
$$

After verification, $f$ satisfies the conditions. Hence $f(2015)=1$.
If $f(-1)=-1$, then $f(y+2)=-f(y)$.
By condition (1), we know
$$
f(x)=\left\{\begin{array}{ll}
0, & x \text { is even; } \\
1, & x \equiv 1(\bmod 4) ; \\
-1, & x \equiv-1(\bmod 4) .
\end{array}\right.
$$

After verification, $f$ satisfies the conditions. Hence $f(2015)=-1$. In summary, $f(2015)= \pm 1$.