AoPS Academy
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
+5 w, let 
be all the positive integers smaller than that are coprime to . Find all 
such that![\[gcd(N, c_i + c_{i + 1}) \neq 1\]](http://latex.artofproblemsolving.com/1/0/7/1075290d442e8c1f54237aad8ad07126f455f1f1.png)
for all 
.
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to .
, 
, 
, 
of positive integers such that for every central sequence 
, 
, 
, , there are infinitely many positive integers with 
.
+2 w
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
be a prime divisor of 
. Notice that 
for any positive integer 
that is not a multiple of 
, we get that either is a multiple of , in which case 
, or 
.
, and thus 
. In particular, it would imply 
, giving the solution 
.. Hence, all 
give 
. Either 
is the only such prime, or is even. In the latter case, 
. So, by a similar logic to above, unless , the only possibility is that is a power of two. In this case, if 
, then the left hand side is 
, which is a contradiction. Hence, again the only possibility is .
. Suppose is even, and let 
Note that for 
, we have 
. Therefore 
. Now consider the LHS mod 
. As 
, for all 
we have![\[a ^ n \equiv \begin{cases}
0 \pmod {2^{k+1}} & \text{if } 2 \mid a \\
1 \pmod {2^{k+1}} & \text{otherwise.}
\end{cases}\]](http://latex.artofproblemsolving.com/9/3/a/93acfff5bc3e7dfd4c5f59eecae5d329e20a7612.png)
Therefore,![\[1^n + 2^n + 3^n + \cdots + n^n \equiv \frac{n}{2} \not\equiv 0 \pmod {2^{k+1}},\]](http://latex.artofproblemsolving.com/9/7/0/9706fe8e82d8b582b4e63e358393b6bead716741.png)
is odd. Consider LHS mod 
. We notice that![\[\left(\frac{n-1}{2}\right)^n + 1 \equiv x! \equiv 0 \pmod {n-1}.\]](http://latex.artofproblemsolving.com/7/4/8/748703dd1e76d0b3fab6b09d01aaedfa91766c13.png)
, we know![\[\frac{n-1}{2} \Biggm| \left(\frac{n-1}{2}\right)^n + 1 \implies \frac{n-1}{2} \mid 1.\]](http://latex.artofproblemsolving.com/1/6/3/1630826d15c225d4ae03781ab4c278cd392b6e2f.png)
, which does not work. Therefore, we have 
. Checking each case by hand we see that 
gives 
and hence we are done. 
n is even
n is odd
is the only odd satisfying given condition
Then mod 
yields, by Euler's Theorem, 
So 
contradiction. Hence is odd. But if 
so by pairing 
contradiction. So 
and testing yields the only solution 
though?
but for me it was motivated by sum of powers and pairing to cancel things out, as is odd.
though? but for me it was motivated by sum of powers and pairing to cancel things out, as is odd.
such that 
