[/quote]
,
Prove that
be a quadrilateral with 
.
the midpoint of the side 
,
.
be the foot of the perpendicular from the point 
.
meets 
at 
,
meets 
at 
.
and 
.
Prove that
,
be positive real numbers, such that 
.![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
,
,
,
lie on a semicircle 
in this order with diameter 
,
.
and 
lie on segments 
and 
.
passes through 
.
,
are concurrent.
,
,
around a circle so that the sum of every 
a grid of 9*9, we choose a little square in 
and its prime fatorization 
,
Prove that there exist infinitely many naturals 
.
that satisfy![\[f(xf(y) + f(x)) = f(x)f(y) + 2f(x) + f(y) - 1,\]](http://latex.artofproblemsolving.com/4/7/f/47f094ce57fe35d339ec3b93152ec3459a22ec48.png)
for every 
,
for every 
and 
,
.
be a set of 
points in a plane that lie within a disk of radius 
billion. Define a 
as picking a point 
and reflecting it across 
moves after which all points of 
such that for every real number 
,
is a perfect square if and only if 
is a perfect square. Prove that 
is the square of a polynomial with real coefficients.
be positive integers such that 
and 
Prove that if 
and 
have the same prime divisors, then 
is a power of 2.
be non-empty subsets of the positive integers, and let 
denote their union. Prove that![\[\sum_{i=1}^m \sum_{(a,b)\in H_i^2}\gcd(a,b)\ge\frac1m \sum_{(a,b)\in S^2}\gcd(a,b).\]](http://latex.artofproblemsolving.com/1/c/d/1cdf561d3af92f524628e0724e0c5ee24d8d6a09.png)
Proposed by Dávid Matolcsi, Berkeley
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
,
as 
,
as 
.
Y by cubres
Let 
be a non-constant monic polynomial with integer coefficients and let 
be an infinite sequence of positive integers. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
.
be a non-constant monic polynomial with integer coefficients and let 
be an infinite sequence of positive integers. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
.This post has been edited 1 time. Last edited by Assassino9931, May 3, 2025, 4:59 PM
is the set of prime divisors of 
.
is the smallest positive integer such that 
and 
,
,
.
.
and if 
is even, 
would give a contradiction. For odd 
thus, 
but then 
where 
which results in a contradiction.
.
,
and 
for 
positive integer. Pick 
,
in order to get 
.
.
for 
.![\[\nu_{p_i}(P(n)^{a_n}+1)\leq \nu_{p_i}((P(n)^{d_i})^{p_1^{\beta_1}\dots p_k^{\beta_k}m}+1)=\nu_{p_i}(P(n)^{d_i}+1)+\beta_i=\nu_{p_i}(P(1)^{d_i}+1)+\beta_i=\beta_i+\theta_i\]](http://latex.artofproblemsolving.com/0/e/9/0e9006891526f624edb7a43227857ce30e9c377d.png)
![\[P(n)^{p_1^{\beta}\dots p_k^{\beta_k}m}+1\leq p_1^{\beta_1+\theta_1}\dots p_k^{\beta_k+\theta_k}=(p_1^{\beta_1}\dots p_k^{\beta_k})(p_1^{\theta_1}\dots p_k^{\theta_k})\]](http://latex.artofproblemsolving.com/3/6/9/3691d3b066d0bf4883fba1fe852756485f6ccabf.png)
Let 
and note that 
is a constant. We have 
.
is unbounded above. 
exists hence 
does not hold. Hence we get contradiction as desired
which is contradiction, good problem.
'
or like the situation when 
?
for some 
,
is unbounded and when it is bounded.
,
is a prime such that 
,![\[ p^{\alpha} \leq (a^{p-1} - 1)p^{\nu_p(k)}. \]](http://latex.artofproblemsolving.com/3/b/a/3baa4f1c3cc466144094c5d7f3634e3987c118bd.png)
,
.
.
is the order of 
modulo 
,
,
.
,
,
is finite, and let 
be all such primes. From the lemma it follows that![\[
P(k)^{a_k} + 1 = \prod_{i=1}^s p_i^{\nu_{p_i}(P(k)^{a_k} + 1)} \leq \prod_{i=1}^s \left(P(k)^{p_i - 1} - 1\right)p_i^{\nu_{p_i}(a_k)} \leq a_k P(k)^{\sum_{i=1}^s p_i},
\]](http://latex.artofproblemsolving.com/3/2/a/32a5c2958d1e21c09a96bf9cc40efd02074b1888.png)
and thus 
.
for all sufficiently large 
,
.
.
for some 
,
,
.
modulo 
,
,
.
,
,
is bounded.
is not a power of 2 for 
because of modulo 4. Therefore, if 
for some 
,
dividing 
,
,
for all 
,
be such that 
for all 
be such that 
,
be sufficiently large. Then for![\[
k = \prod_{i=1}^s p_i^{2N} + t
\]](http://latex.artofproblemsolving.com/8/6/8/8685b6dcd274fb3ff18d35fcbee09b5b9955936d.png)
we have![\[
P(k)^{a_k} + 1 \equiv P(t)^{a_k} + 1 \pmod{p_1^N p_2^N \cdots p_s^N}.
\]](http://latex.artofproblemsolving.com/0/4/d/04de40f81ff5c3a0f55404b6a541255a33eef7fe.png)
Therefore, 
,![\[
P(k)^{a_k} + 1 \leq \prod_{i=1}^s p_i^N < \sqrt{k},
\]](http://latex.artofproblemsolving.com/7/5/d/75d24ffa7400081cdbe9b7109a9dbdc91e903e49.png)
which leads to a contradiction for sufficiently large 
.
for some 
,
,
such that each natural number appears in the sequence no more than 
are a finite number of nonconstant polynomials, and each of them attains a given value only finitely many times; here 
be all the primes with the required property. Write 
for 
.
for every 
,![\[
b_k < c^{k^{\varepsilon}} = 2^{k^{\varepsilon} \log_2 c}
\]](http://latex.artofproblemsolving.com/f/2/d/f2df46178ee2ec6378a6d2b828099c275af0e47b.png)
for large 
does not exceed 
.
does not exceed 
.
.![\[
m^{1 - s\varepsilon} \leq T (\log_2 c)^s.
\]](http://latex.artofproblemsolving.com/c/2/a/c2a77e687bc15caf815a04e79940d92eb365625d.png)
If we choose 
,
,
