AoPS Academy
with incircle 
. Let 
be the point on such that the circumcircle of 
is tangent to and let the 
-excircle be tangent to 
at 
. Prove that the tangent from to and the tangent from to (distinct from ) meet on the line parallel to and passing through .
by 
, 
for every positive integer 
. Prove that 
leave distinct remainders when divided by 
.
+2 w
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.
satisfies the following condition., ![$\sum_{k=1}^{n}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n}{k}\right]}\right)a_k=1$](http://latex.artofproblemsolving.com/3/1/0/3109e3f024dc59c2cdad60efa270494d6095c330.png)
holds.
, compute 
.
, too.
![$$\sum_{k=1}^{n}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n}{k}\right]}\right)a_k=1=\sum_{k=1}^{n-1}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n-1}{k}\right]}\right)a_k \implies \sum_{k=1}^{n}\left((-1)^{\left[\frac{n-1}{k}\right]} - (-1)^{\left[\frac{n}{k}\right]}\right)a_k=0$$](http://latex.artofproblemsolving.com/0/8/b/08bb6a81eb96ae710107732001e2565f1819b0db.png)
![$$\left(\left[\frac{n}{k}\right]\ne \left[\frac{n-1}{k}\right]\iff k|n\right) \implies \sum_{d|n}(-1)^{\frac{n}{d}}a_d=0$$](http://latex.artofproblemsolving.com/7/5/9/75982029aa62b9de18cc360d686f9ccd3db2c667.png)
so if is a power of 
then 
.
Proof: Let 
where 
is odd.
. Assume that 
, then 
and this proves the claim.
By the mobius inversion formula 
so
And the proof is complete.
and 
.
.![$\sum_{k=1}^{n}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n}{k}\right]}\right)a_k - \sum_{k=1}^{n-1}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n-1}{k}\right]}\right)a_k$](http://latex.artofproblemsolving.com/f/c/1/fc1197226f6f8f124648e95074d8fab5bc2258f5.png)
.
