be a triangle with circumcircle 
and incentre 
.
intersects the lines 
,
,
at points 
,
,
,
,
,
,
,
,
of the segments 
,
,
,
.
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
,
.
![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
denote the set of single-variable polynomials in ![$\theta : \mathbb{Z}[x] \to \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/3/2/032a637d4ae21da2c0ddd8dc127723807f11cc5f.png)
(i.e. functions taking polynomials to polynomials)![$p, q \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/7/e/7/7e7eb925258a33e456ef4ff722c1f943dc155d10.png)
,
;![$p \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/2/6/0267b98c16b400dda1f081877e9b84fe672ce0e7.png)
,
has an integer root if and only if 
does.
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
.
such that no sum of three distinct elements of 
.
.
that satisfy the equation 
.
be a circle, and let a line 
and 
Circles 
and 
are internally tangent to 
and 
respectively, and both are tangent to 
Similarly, circles 
and 
are externally tangent to 
respectively.
the segment 
has a length of 
and 
find the length of segment 
and 
Prove that
.
Prove that
and 
Prove that
Y by Orestis_Lignos, rama1728, Pitagar, samrocksnature, jasperE3, megarnie
Let 
be any positive integer. You are given these two quantities:
You are allowed to apply any of these four operations:
from the current one by a finite (possibly empty) sequence of such operations.
be any positive integer. You are given these two quantities:
You are allowed to apply any of these four operations:
- You can add or subtract any positive integer onto/from the numerators of both the quantities (simultaneously)
- You can add or subtract any positive integer onto/from the denominators of both the quantities (simultaneously)
- You can reduce any quantity to it's lowest form (You need not reduce both quantities simultaneously)
- You can interchange the position of the two quantites
from the current one by a finite (possibly empty) sequence of such operations.
and 
you can attain 
and 
for every 
as follows:
,
and 
in order of written in the problem paper(4 is almost not neccesary, that's why I've ignored it).
you proceed as follows:
Note that on the 3rd last operation 
,
some of the steps may be skipped or just are nonexistent, but that does not affect the solution.
you proceed as follows:
Note that on the 3rd last operation 
,
are, at least one of 
.
.
such that
.
.
.
denote addition onto the numerators of both quantities, let 
denote addition onto the denominators of both quantities, denote 
as reduction of right quantity to its lowest form and 
as reduction of left quantity to its lowest form.
,
.
Furthermore we reduce our two quantites to
Now, we show that we can do this sequence backwards, i.e. we get from 
to any 
.
Obviously, number of such operations is finite, thus we are done.
to the numerator, we get 
to the numerators, we get 
.
is allowed
,
is positive, and so 
are positive. Also, 
,
and 
,
sign between the fractions instead of 
but it won't matter.
.
be any divisor of 
.
from 
.
.
then 
.
where 
is some divisor of 
.
to be any divisor of 
for some 
.
:
.
if 
.
then one can obtain 
from 
for any 
:
.
then you can reach 
from 
where 
is some positive integer depending on 
.
,
,
.
and 
,
.
and change the denominator of the second fraction to 
(as 
,
and the denominator 
.
and decreases the denominator by 
,
and changes the denominator difference to 
.
and 
as these two fractions have the same numerator and denominator differences as the current state. From here, simplify the first fraction to get 
and 
In general from the last line we can achieve any 
for 
.
Since this can be done for 
we can get 
for all 
.
Now notice that all the steps here are legal when 
so we can get 
for 
but at the beggining we got it for 
and then for 
therefore we can achieve such pair for any positive integer 
.
