be an odd prime and 
be integers so that the integers 
are divisible by 
such that:
for any positive integer 
,
are all distinct, and
,
?
lies on a line segment 
between 
and 
and circles are drawn having 
and 
as diameters. A common tangent to both circles touches the circle with 
and the circle with 
.
and the common tangent to both circles at 
and 
denote three distinct integers, and let 
denote a polynomial having integer coefficients. Show that it is impossible that 
and 
.
Y by
Let 
be a triangle whose incircle is tangent to 
, 
, 
at 
, respectively. Let 
and 
be the midpoints of 
and 
, respectively. Suppose that points 
and 
lie on 
so that 
and 
. Prove that lines 
and 
intersect on the circumcircle of 
.
be a triangle whose incircle is tangent to 
, 
, 
at 
, respectively. Let 
and 
be the midpoints of 
and 
, respectively. Suppose that points 
and 
lie on 
so that 
and 
. Prove that lines 
and 
intersect on the circumcircle of 
.
.
,
is actually the midpoint of 
.
and 
are collinear. Therefore![\[\angle UBC = \angle UBN + \angle NBC = \angle BNU + \angle IBC = \angle INM + \angle IBC = \angle IDM + \angle IBC = \angle BCI + \angle IBC.\]](http://latex.artofproblemsolving.com/6/a/3/6a389f01973361dc72ae128c44b8dcd7d477ad89.png)
This gives 
,
meet at midpoint of arc 
.
meet 
again at a point 
.
cyclic and![\[\angle LBI = \angle UNB = \angle INM = \angle IDM = \angle ICB = \frac 12 \angle C.\]](http://latex.artofproblemsolving.com/7/2/2/7220fb8ba407bc0c5632294d3a94f01c44f19f99.png)
Thus 
.
,
gives the desired concurrency on 
be the incenter of 
where we use the fact that 
is tangent to 
.
.
.
where we use the fact that 
is cyclic. Thus, we get that 
.
so 
,
because why not. 
Done.
