are such that the numer
is rational. Find all possible values of 
or show that there are infinitely many.
be a positive integer that is not divisible by 
.
be a positive integer. Prove that none of the numbers of the form 
is a square number.
intersect at point 
.
and 
with their common external tangents lie on a circle 
.
and 
with their common external tangents lie on a circle 
.
be a positive integer. Furthermore, let ![$x_1, x_2,\ldots, x_n \in [0, 2]$](http://latex.artofproblemsolving.com/b/2/a/b2a6731961601353e26224c285a32923fbb54eb3.png)
be real numbers subject to 
.
When does equality hold?
be a real number such that the system 
has exactly one complex solution 
The sum of all possible values of 
where 
and 
Here 
Y by
Find all integer coefficient polynomial 
such that for all positive integer 
, we have 
Where 
denotes the number of divisors of 
. Define 
.
Note: you can use this conclusion. For all
, there exists a positive constant 
such that for all positive integer , the th smallest prime is at most 
.
Proposed by USJL
such that for all positive integer 
, we have 
Where 
denotes the number of divisors of 
. Define 
.Note: you can use this conclusion. For all
, there exists a positive constant 
such that for all positive integer , the th smallest prime is at most 
.Proposed by USJL
This post has been edited 1 time. Last edited by CrazyInMath, Sep 11, 2022, 12:21 PM
has non zero constant term 
Let 
be any prime number. Put 
where 
be the sequence of the prime numbers and 
Note that 
,
or greater or equal to 
.
is small. There must be a lot of large prime factors (with their multiplicity) in order to attain 
divisors, so when 
is bigger than 1.
,
,
denote the number of not necessary dinstinct prime factors of 
that lies in the interval ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
.
to be the k-th prime number and 
,
.
such that 
.
.
,
behind but there are 
primes, therefore I don't see any reason it does not work here.
then 
is determined by 
.
then 
so 
.
isn't a solution. Assuming the contrary. Let 
and 
is a positive constant such that 
holds for all 
.
be a positive integer large enough with respect to 
be the prime sequence and 
be the set of the first 
of ![\[x_I=\prod_{p\in S\backslash I}p.\]](http://latex.artofproblemsolving.com/1/6/e/16eb3a5cdd0490d6688708c37a35b362727f1e8d.png)
Then 
.
.
be a positive integer yet to be determined. then![\[\tau(P(x_I))=\prod_{p\leq M, p|P(x_i)}(v_p(P(x_I))+1)\prod_{p> M, p|P(x_i)}(v_p(P(x_I))+1)\]](http://latex.artofproblemsolving.com/e/d/4/ed4de09ad768d987a43f18ffe31c3d42f968fc66.png)
![\[\leq \left(\log_2 Cp_N^{n(N-d)}\right)^{\#\{p|p\leq M, p|P(x_I)\}}\cdot 2^{\log MCp_N^{n(N-d)}}.\]](http://latex.artofproblemsolving.com/9/f/b/9fb50e391986b69a194ebc629774270d799211af.png)
Now we take every subset 
,
to hold.
,
that is at most 
contributes at most 
.
has at most 
.
,![$(2^{t-1}p_N, 2^tp_N]$](http://latex.artofproblemsolving.com/1/8/0/1801fa16c29cf48fe46b24b75072eb22eefeb30f.png)
is at most![\[2^tp_N\cdot \frac{2p_N^d}{2^{t-1}p_N}=4p_N^d.\]](http://latex.artofproblemsolving.com/4/7/0/47034140dcefb479ef1f1cefa5b5381ac425ab98.png)
Take 
,
be the largest integer satisfying ![\[\prod_{I\in\binom{S}{d}}\tau(P(x_I))\leq \left(\log_2 Cp_N^{n(N-d)}\right)^{(k+d)\binom{N}{d}+4tp_N^d}\cdot 2^{\binom{N}{d}\log_{p_N^d}Cp_N^{n(N-d)}}.\]](http://latex.artofproblemsolving.com/3/d/0/3d0b4b5d057eb326f47022f13def09195eb38c88.png)
However,![\[\prod_{I\in\binom{S}{d}}\tau(P(x_I))\geq 2^{(N-d)\binom{N}{d}}.\]](http://latex.artofproblemsolving.com/b/4/5/b459442088eafef8fe9a976e3b004f36926dbbb0.png)
Take log with base 
,
and leading coefficient 
.
adding some polynomials at most degree 
.
and 
