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be an integer greater than 
, and let 
be non-negative real numbers with 
. Determine the minimum value of ![\[ \frac{a_1}{a_2^2 + 1}+ \frac{a_2}{a^2_3 + 1}+ \cdots + \frac{a_n}{a^2_1 + 1}.\]](http://latex.artofproblemsolving.com/1/3/7/1374198f13a8e0a8c3507c5f086c69ad3a719b3b.png)
, 
, 
.
for 
and suppose 
. Prove that for any positive integer 
,![\[ \left( a_1^k + \frac{1}{a_1^k} \right) \left( a_2^k + \frac{1}{a_2^k} \right) \dots \left( a_n^k + \frac{1}{a_n^k} \right) \ge \left( n^k + \frac{1}{n^k} \right)^n. \]](http://latex.artofproblemsolving.com/5/a/d/5ad4ea89463088b023d252d18d9f6eb411ec9f64.png)
is selected on arc 
not containing 
on 
. 
and 
are the reflections of point 
and 
with respect to 
, respectively. Circumcircles of 
and 
intersect at 
. Prove that 
is colinear
we see that all prime factors of 
is 
. It is well known that there are infinitely many primes 
such that there exist 
satisfying 
if 
is non-constant. For any such , we can assume 
but it follows that 
. So for all such one has that 
where 
is the constant coefficient. Then 
and we can just consider 
which satisfies the same condition as . From this we get that 
for some integer 
and and it follows that 
or 
from checking 
and that it sufficies to have 
by checking 
.
. Now suppose the polynomial is non-constant. Now we prove a lemma:, we have 
.
. The base couple cases are trivial by exhausting the few possibilities. Now it suffices to check all 
. Now take some 
. Then take 
where 
. After that it is clear that 
so we are done.
. Dividing through by we receive a new function satisfying the same requirements as , implying that our only solutions are of the form 
for some non-negative 
. 
for 
. To see that these work, note that 
for all positive integer 
.
is constant it must be 
, so assume otherwise. I claim only prime factors of may divide . Note that 
for integer polynomials, so let 
be equivalent to modulo some prime such that 
and thus 
. But 
so it is not divisible by , so is not divisible by any prime that does not divide and the claim is proven.
for some 
, but as is nonconstant, we have 
for infinitely many primes . As obviously divides all the nonconstant terms of 
, we then know that the constant term of is divisible by infinitely many primes, and thus 's constant term is zero. But then we can factor an out of and our arguments still hold, so by induction we get that is and we are done.
.This means that 
.Then let 
be the set of prime divisors of .If 
then 
.Write 
.Using similar arguments , we will get that 
.Thus , we can reduce the degree and ultimately get 
Since 
, we have that 
.Also 
where 
and 
was constant then it would be 
or by plugging 
. So now assume that its not constant, hence by Schur we have that we can set a larger enough prime such that for some we have 
and now we can set that 
and 
to get 
which would mean that 
hence 
but aince we said we can let as larger as we want we have that 
hence 
which means that the replace 
makes sense, now from here its very easy that we can repeat the same process until we get 
in which here 
has to be or hence we are done 
**
(where 
). Obviously they work. Now we will prove the converse. Assume is nonconstant (otherwise only solution is 
obviously). Define ![\[f(X):=a_dX^d + \dots + a_0\]](http://latex.artofproblemsolving.com/7/1/2/712194cdef863cd406820890051ed0f651cebb11.png)
Now see that by Schur's Theorem there exists infinitely many primes such that for some . Consider the minimal such .
or else 
contradicting minimality. But also 
, and hence and we get ![\[p \mid f(p) \iff p \mid f(0) \text{ for infinitely many primes $p$ } \iff f(0)=0 \iff a_0=0\]](http://latex.artofproblemsolving.com/a/4/e/a4ec9d539738837a41659c6987984ca928db153f.png)
and hence now we can define ![\[g(X):=\frac{f(X)}{X}\]](http://latex.artofproblemsolving.com/6/a/0/6a0cb7dc7e07ce290a88e51b4a7882a38d37d0c6.png)
and repeat the process untill we get 
.
and 
, we get the above solutions, as desired.
, determine all polynomials with integer coefficients such that divides 
for every positive integer .
be a prime factor of , with 
fixed 
(
is the maximum that satisfies)
for 
. are a subset of the prime factors of (without multiplicity).
. For 
then all prime factors of are 
, and for each we have 
, so 
as desired. is a positive or negative power of with nonnegative exponent 
for all . Thus there exist some 
for which 
for infinitely many . Both sides are polynomial in , so since they are equal infinitely many times they are the same polynomial, as desired.
where 
It follows that 
If 
then 
which is a contradiction since 
where the last value has no factor of 
then 
Thus, 
Assume 
It follows that 
which forces 
There are infinitely many such that either 
and then subtracting/adding 1 implies our result. such that 
for some 
This implies the result.
because 
.
,
so we have contradiction.
be sum of coefients with odd deg (e.g 
)
similarly.
we have 
so 
or 
so putting 
.
so 
so puting 
.
where 
and 
goes from 
to 
