AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a triangle with circumcircle 
and let 
be the 
-excircle. Let 
and 
be the intersection points of and . Let 
and 
be the projections of onto the tangent lines to at and respectively. The tangent line at to the circumcircle of the triangle 
intersects the tangent line at to the circumcircle of the triangle 
at a point 
. Prove that 
.
+1 w
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
is lovely if there exist a positive integer 
and (not necessarily distinct) positive integers 
, 
, 
, 
such that 
and 
for 
.
, which is a perfect square of an integer?
, prove that there exists a positive integer 
, such that for any integer 
, can be expressed by a sum of positive integers 
's as![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
, 
, 
, 
.
be a function such that
for all 
. Determine all such functions 
.
such that the equality 
holds for all pairs of real numbers 
. with integer coefficients such that 
and 
is a square of an integer for all nonnegative integers 
. and an integer , 
is defined as follows: 
if 
and 
if 
.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
. be a positive integer. A positive integer is called a benefactor of if the positive divisors of can be partitioned into two sets and 
such that is equal to the sum of elements in minus the sum of the elements in . Note that or could be empty, and that the sum of the elements of the empty set is 
.
is a benefactor of 
because 
. has at least 
benefactors.
denotes the least natural such that
Find all naturals 
such that 
.
are 
, 
, 
, 
, and 
. For 
, we take the following cases:
, take 
.
, take 
.
, let 
be the smaller solution in 
to 
, which exists since the discriminant is 
and is a quadratic residue mod 
. It is not hard to check that 
, so we can take 
.
for which![\[\left\{2^{-1}, 3^{-1}, \ldots, \left(\frac{p-1}{2}\right)^{-1}\right\} = \left\{-2, -3, \ldots, -\frac{p-1}{2}\right\} \pmod{p}.\]](http://latex.artofproblemsolving.com/e/a/b/eab1eb56436a33d87c2c5f77c799a7ee27dab26d.png)
which are easily checkable. Now we prove that any other prime is not mundane. Note that 
is clearly not mundane because of 
. Let 
.
then since , we can set 
and 
.
then since , we can set and 
.
then since , we can set 
and 
.
then since , we can set 
and 
, which works because 
and 
.
and 
: Prove that :
![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
is an obvious solution. Now, let 
.
or 
.
.
such that 
is a positive integer. Find, with proof, all prime numbers that are not mundane.
check by hand that only 
and 
,![\[(a,b)=\begin{cases}
\left(\frac{2p-2}{5},\frac{p-5}{2}\right) & p \equiv 1 \pmod{5} \\
\left(\frac{2p+1}{5},5\right) & p \equiv 2 \pmod{5} \\
\left(\frac{3p+1}{10},10\right) & p \equiv 3 \pmod{5} \\
\left(\frac{p+1}{5},5\right) & p \equiv 4 \pmod{5} \\
\end{cases} \]](http://latex.artofproblemsolving.com/f/3/1/f31788496b757f35a9bad72566972493ef380dbf.png)
,
manually and assume 
.![\[\left\{-10, -\frac{p-5}{2}\right\} \in \left\{-2, -3, \dots, -\frac{p-1}{2}\right\},\]](http://latex.artofproblemsolving.com/d/c/f/dcf0ab3cdd34ff1d8ac4fd904e0b970970c70812.png)
but their reciprocals differ by![\[\frac{1}{10} - \frac{2}{p-5} = \frac{p-1}{2}\]](http://latex.artofproblemsolving.com/6/f/6/6f61f9ea21739c13b747e833f5f3ecab9889c7c2.png)
and thus cannot both lie in 
.
manually, and assume 
for some odd 
,
.
must be prime, both 
and 
are consecutive Fermat primes. Since 
and 
,
.
,
.
must be prime, 
.
gives 
.
,![\[5 \mid 3 \cdot 2^{2^c-1} + 1 = p,\]](http://latex.artofproblemsolving.com/e/b/b/ebb17463b469c0e55e019a6233144705760eef73.png)
contradiction.
,![\[2^{k+1} \cdot \frac{p-n}{2} \equiv 1 \pmod{p}\]](http://latex.artofproblemsolving.com/7/f/a/7fa8f0447c3bf1c5d56a7c306bed42096d161011.png)
shows that this case yields no solutions.
be the smallest prime not dividing 
.
unless 
.
.
gives 
gives 
.
gives 
.
gives 
.![\[13^2 < \frac{2 \cdot 3 \cdot 5 \cdot 7 \cdot 11}{2}\]](http://latex.artofproblemsolving.com/5/b/9/5b98f220ce91778cb6e4e0299ef92216f6cb834b.png)
and 
for 
,
are the primes in increasing order. 
by minimality of 
,
.
,![\[\left(q \operatorname{rad}(k+1)\right) \cdot \frac{kp+1}{q \operatorname{rad}(k+1)} \equiv 1 \pmod{p}.\]](http://latex.artofproblemsolving.com/b/8/1/b81d2bfc75bfd2ab6b740cd18a5a7190a66e85eb.png)
If 
,![\[q \operatorname{rad}(k+1) < q^2 < \frac{p}{2}\]](http://latex.artofproblemsolving.com/8/6/d/86ded885ddc652c06e767ca1d78c18bd6ce54544.png)
and the second factor as![\[\frac{kp+1}{q \operatorname{rad}(k+1)} < \frac{(k+1)p}{(k+1)2} = \frac{p}{2}.\]](http://latex.artofproblemsolving.com/4/9/0/49077cf35734a27e1058e3e5db137b02863aa69e.png)
Therefore, no 
work. Indeed,
as desired.
.
if 
and 
otherwise. Thus, a prime 
for which 
.
are not mundane.
,
.
,
,
.
.
,
,
.
,
,
.
,
,
.
for which 
for some positive integer 
,
,
are divisible by 
.
,
are divisible by 
,
,
,
is not mundane.
,
,
.
are mundane.
which was not mundane (so, from now on, when we refer to 
,
.
.
,
.
.
.
,
modulo 
as 
,
.
,
as well 
,
it follows that 
be the unique positive integer for which 
.
,
,
,
.
,
is a positive integer, 
,
contradicting the fact that 
,