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, 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.
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
denote the set of positive integers. Find all functions 
such that for positive integers and 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
and 
are erected outside an acute triangle 
Suppose that ![\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]](http://latex.artofproblemsolving.com/3/3/7/3378f0c4409d14bc0afddf2616992c13f624be3b.png)
Prove that lines 
and 
are concurrent.
. 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 
are the roots of the cubic equation 
then compute 
.
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