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.
and 
are the roots of the cubic equation 
then compute 
.
be a positive integer that is not divisible by 
. Furthermore, let 
be a positive integer. Prove that none of the numbers of the form 
is a square number.
?
, and the sum of the roots is 
. So we need to find a combination of integers that multiply to and add up to . Let's call the roots of the equation p, q, and r. From Vieta's, we get that 
and 
. Looking at the factors of , which is 
, we try to split the numbers in a way that gives us the correct sum and product. Trying 3, -2, and -8, we see that they add up to 
and multiply to 
, so they work. That means the roots of the polynomial are -3, -2, and -8, and the factorization is 
. Multiplying it out, we get 
, so we find that a = 46.
, 
, 
, 
lie on a semicircle 
in this order with diameter 
, and is not parallel to 
. Points 
and 
lie on segments 
and 
respectively such that 
and 
. A circle 
passes through and is tangent to , and intersects again at 
. Prove that the lines 
, and 
are concurrent.
if the following conditions are satisfied: 
, 
, 
.
be a triangle with incircle touching 
at 
respectively. Let 
and 
be its circumcenter and midpoint of 
Suppose that circumcircles of 
and intersect at for the second time. Assume 
is on the circumcircle of such that 
is cyclic. Prove that circumcenter of 
lies on 
satisfy 
. The circumcircle 
and the incircle 
of are tangent to the sides 
at 
, respectively. The line 
meets 
at and intersects at 
, while the line 
meets 
at 
. Construct the common external tangent 
(different from ) to the incircles of the triangles 
and 
. Show that is parallel to the line 
.
are being replaced with 
. Find the biggest number of steps, after which the number 
will appear on the board.
.
the equation: 
, 
. Points 
lie on segments 
respectively. Points 
are such that 
and 
are equlateral, and 
lie on opposite sides of 
, and 
lie on the opposite siedes of . Prove that
is a parallelogram. Take 
on the line 
so that 
and lies between and . Let the line through perpendicular to and the line through perpendicular to meet at 
. Show that 
.
. How many complex numbers are there such that 
and 
is an integer.
then 
and 
. Note that 
and we can only have equality when 
. Then 
. This 
gives 120 solutions for , from now on 
.
then 
hence 
where 
. If 
then 
impossible. If 
then consider 
. We have 
and 
. This is impossible as we just saw.
gives no solutions. The final case is 
hence 
. This gives 600 solutions. Hence in total we have 720 solutions.
and 
are complex numbers such that
.
and
Since this quantity is an integer, it must be real, so 
From here we have two cases.
(side note: is it an abuse of notation to use "mod" for non-integer values?), or, put simply, 
and 
are the same angle. In this case, 
is already 
,
possible values for 
or that 
Rearranging, we get that this is equivalent to 
which also means that 
so 
is an integer if and only if 
is. From here, we have several cases:
Hence, the only 
,
total possible 
different 
)
for 
.
such