AoPS Academy
such that 
. Show that the equation 
has no integer solution.
denotes a polynomial of degree 
such that 
for 
, determine 
.
with 
, the foot of altitudes from 
to the sides 
are 
, respectively. 
is the orthocenter. 
is the midpoint of segment 
. Lines 
and 
intersect at 
. Let the tangents drawn to circumcircle 
from 
and 
intersect at 
. Prove that 
are colinear
![$\ [\text{Blog Post 60}] $](http://latex.artofproblemsolving.com/2/a/a/2aae5ba574706ab8747e64cb5ac6ef9a452fac75.png)
, find the minimum value of 
if
is natural and 
be a prime number. Prove that there exists a natural number 
such that![\[
p \mid m^n - n.
\]](http://latex.artofproblemsolving.com/3/a/3/3a38b299f0328d176286d65e02b92e9918e8109d.png)
just pick 
, otherwise 
is coprime to 
. By Chinese Remainder Theorem there exists an 
such that 
and 
, pick this , we see that both 
and leave remainder 
modulo , so we are done.
![\[ \frac{d}{dx}\left[\dfrac{4x^2+8x+13}{6(1+x)}\right]= \dfrac{(6+6x)(8+8x)-(4x^2+8x+13)(6)}{(6+6x)^2} = \dfrac{24x^2+48x-30}{(6+6x)^2}=\dfrac{4x^2+8x-5}{x+1} \]](http://latex.artofproblemsolving.com/5/b/7/5b72aee56d9a0cca6cdcb313d225be89fec3cee7.png)
![\[ \dfrac{4x^2+8x-5}{x+1}=0 \]](http://latex.artofproblemsolving.com/0/b/c/0bca2ae77fb5cc16c5ee164daad4bd0e59d07353.png)
cannot equal 
,
being a possibility.![\[ 4x^2+8x-5 = 0 \]](http://latex.artofproblemsolving.com/3/2/f/32f34881ae80e44a8346157a9d121f2890c062f5.png)
![\[ (2x-1)(2x+5) = 0 \]](http://latex.artofproblemsolving.com/4/3/d/43de8d4361ae531ad28d33d0ec5841f07ae82862.png)
![\[ x=\dfrac{1}{2},-\dfrac{5}{2} \]](http://latex.artofproblemsolving.com/d/7/6/d763c952015b20c35ca086d6dfe7e5eaa671d1fd.png)
to find what the value of ![\[ f\left(\dfrac{1}{2}\right)= \dfrac{4\left(\dfrac{1}{2}\right)^2+8\left(\dfrac{1}{2}\right)+13}{6\left(1+\left(\dfrac{1}{2}\right)\right)} \]](http://latex.artofproblemsolving.com/b/2/8/b2801124f3b260864cb93f294f8d35bc533001bf.png)
![\[ \dfrac{1+4+13}{6+3} = \dfrac{18}{9} = 2 \]](http://latex.artofproblemsolving.com/1/2/6/126b8823b2a3457a357d6f87cb0c4fb764f3c24c.png)
![\[ \dfrac{d^2y}{dx^2}\left[f(x)\right]=\dfrac{(x+1)(8x+8)-(4x^2+8x-5)(1)}{(x+1)^2} \]](http://latex.artofproblemsolving.com/d/e/4/de476cfe6d3d2f986f3be2d07b89245859eb3259.png)
)
is also a minimum. Try plugging in ![\[ f(x)= \dfrac{4x^2+8x+13}{6(1+x)} = \dfrac{4(x+1)^2+9}{6(x+1)} \]](http://latex.artofproblemsolving.com/c/c/f/ccf872f850f66aac087dc93863d7bda8f6cf4151.png)
![\[ f(x)=\dfrac{2}{3}(x+1)+\dfrac{3}{2(x+1)} \]](http://latex.artofproblemsolving.com/5/9/c/59ca577b4a20d6f5a4d1cecef381d2d1e02288e2.png)
![\[ \dfrac{\dfrac{2}{3}(x+1)+\dfrac{3}{2(x+1)}}{2} \geq \sqrt{\dfrac{2}{3}(x+1)\cdot\dfrac{3}{2(x+1)}} \]](http://latex.artofproblemsolving.com/3/d/2/3d22bb103d8e2875ddbbc7e74383fad776f58cfb.png)
![\[ \dfrac{\dfrac{2}{3}(x+1)+\dfrac{3}{2(x+1)}}{2} \geq \sqrt{1} \]](http://latex.artofproblemsolving.com/6/2/a/62aa71af130a1de8cab94b8de0090f8fb2ba28dc.png)
![\[ \dfrac{3}{2}(x+1)+\dfrac{3}{2(x+1)} \geq 2 \]](http://latex.artofproblemsolving.com/9/f/6/9f69536215ae4039656479fd24e3e8e0eafeedf9.png)
![\[ \therefore f(x)=\dfrac{4x^2+8x+13}{6(1+x)}\geq 2 \]](http://latex.artofproblemsolving.com/5/e/b/5eb01c18e06645862ef5a3a6b0391dac92cec464.png)
,
