Sequence modulo 2011

A sequence of integers $a_1, \dots, a_{2010}$ satisfy the following properties:

$a_1 - 1$ is divisible by 2011
$a_k a_{k-1} - k$ is divisible by 2011 for $k = 2, 3, \dots, 2010$

Show that $a_{2010} + 1$ is divisible by 2011.

Solution

First we prove the following assertion:

For $k$ odd, then $a_k.2.4.6. \cdots.(k-1) - 1.3.5.\cdots.k$ is divisible by 2011.
For $k$ even, then $a_k.1.3.5. \cdots .(k-1) - 2.4.6.\cdots.k$ is divisible by 2011.

Proof by induction. It can easily be seen for $k=1$ it's true. For $k=2$, we have $2011 | a_2a_1 - 2 = a_2(a_1-1) + (a_2-2)$ So $2011 | a_2 - 2$

Suppose it's true for $k-1$ odd, then for $k$ even:
$2011 | a_k a_{k-1} - k$ which means
$$2011 | 2.4.\cdots.(k-2) (a_k a_{k-1} - k) = 2.4.\cdots.(k-2)a_k a_{k-1} - 2.4.\cdots.(k-2).k$$
$$2011 | a_k (2.\cdots.(k-2)a_{k-1} - 1.3.\cdots.(k-1)) + (a_k.1.3.\cdots.(k-1) - 2.\cdots.(k-2).k)$$
And since 2011 divides the first term by induction hypothesis, then it also divides the second term, which completes the induction step. The proof for $k-1$ even and $k$ odd is similar.

Now, that means for $k=2010$, let $X = a_{2010}$ we have:
$$2011 | X.1.3. \cdots .2009 - 2.4.\cdots.2010$$

We now prove the following assertion for $i = 1,2,\dots,1005$
$$2011 | X.1.3. \cdots. (2011-2i) + (-1)^i (2i) (2i+2)\cdots.2010$$

For $i = 1$ it is true by definition. Now suppose it's true for $i-1$, then for $i > 1$:
$$2011 | X.1.3. \cdots. (2011 - 2i).(2011-2i+2) +(-1)^{i-1} (2i-2) (2i)\cdots.2010$$
if $2011 | a(2011+b) + c$ then $2011 | ab+c$
$$2011 | X.1.3. \cdots. (2011 - 2i).(-2i+2) + (-1)^{i-1} (2i-2) (2i)\cdots.2010$$
if $2011 | a$ then $2011 | -a$.
$$2011 | X.1.3. \cdots. (2011 - 2i).(2i-2) + (-1)^{i} (2i-2) (2i)\cdots.2010$$
$$2011 | 2(i-1)(X.1.3. \cdots. (2011 - 2i).+ (-1)^{i} (2i)\cdots.2010)$$
if $2011 | 2(i-1)a$ then $2011 | a$ since 2011 is prime and $i > 1$.

So the assertion holds for $i = 1,\dots, 1005$. Particularly, for $i = 1005$ we have:
$$2011 | X - 2010$$
$$2011 | X + 1 - 2011$$
$$2011 | X + 1$$