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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
such that there exist integers 
with ![\[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\]](http://latex.artofproblemsolving.com/e/f/a/efaa98e77ac0273284ee3446e4335d8cc4f681cb.png)
. If 3 and 
are coprime, 
, so 
by Euler's Theorem. Therefore 
or 
. However, 
, so if t<4, 
. Therefore, 
.
, so this must be our minimum achievable value. 
such that there exist integers with 
.
works.
where 
.
is optimal.
works.
.The right side is: 
.
, 
. Hence the sum of any three or two of these won't be equal to 
.
is indeed a really big number, so it would be quite nice if we could somehow reduce it to something feasible, at least something to the order of a thousand... wait this is a no-brainer, let the 
of our numbers be 
, which of course is a perfect cube. Thus let's write 
to get:
.
must satisfy this equation- aha so we have found that indeed works, and thus we've got to prove that no other 
(aka 
and 
) ends up working... well, this seems like an easy prey for some simple modular arithmetic manipulation...
will be quite useless here, so let's start with 
, as we know 
... now let's try to fiind 
:
- thus 
, and as 
, we're done., with 
![\[ \frac{d}{dx}\left[\dfrac{4x^2+8x+13}{6(1+x)}\right]= \dfrac{(6+6x)(8+8x)-(4x^2+8x+13)(6)}{(6+6x)^2} = \dfrac{24x^2+48x-30}{(6+6x)^2}=\dfrac{4x^2+8x-5}{x+1} \]](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)
,
is congruent to 4 (mod9)., we need at least 4 terms.
cubes is attainable 
so 
.
is not satisfies. We have ![\[ 2002^{2002}\equiv 4^{2002}\equiv 4\pmod 9.\]](http://latex.artofproblemsolving.com/5/e/6/5e672ff0642825fc556062006af9ebb4f5642969.png)
Since 
,
for 
and 
,
from Euler's Theorem. Since the set of cubic residues mod 
,
.
so 
we get that![\[x_1^3 + x_2^3 + \cdots + x_t^3 = 2002^{2002} \equiv 4 \pmod{9}\]](http://latex.artofproblemsolving.com/0/3/0/030592aae3cd928965b8458eb339b21232320616.png)
and since the only cubic residues modulo 
,
.
:![\[(2002^{667})^3 + (2002^{667})^3 + (10\cdot 2002^{667})^3 + (10\cdot 2002^{667})^3 = 2002^{2002}\]](http://latex.artofproblemsolving.com/4/c/3/4c3ba5157d0469fa46b02bd79f8aed1a3cde2646.png)
Using 
we get 
so we need at least 
to end the problem.
and cubes are 
in 
,
.
is obtainable because 
.
for all integers 
.
,
.
is attainable; consider 
.
.
,
.
.
,
cubes.
Now, to show that this is optimal, note that taking 
,
.
for all integers 
.
(
,
,
.
which is not possible)
is mentioned by many posts above
Taking the equation 
but since each 
clearly 
Now we construct a solution for 
showing that this is the answer. Note that we can just do 
so this finishes off the problem.
and 
.
and 
:
.
.
Now I prove minimality.
By taking our original equation modulo 9, we get that 
Thus, 