AoPS Academy
, denote the sum of digits of in its decimal representation by 
. Find all polynomials 
with integer coefficients such that for any positive integer 
, the integer 
is positive and 
is called 
if there exists a permutation 
of the numbers 
, such that:
and 
have different parities for every 
;
the sum 
is a quadratic residue modulo 
for every 
.
of positive integers with 
prime and ![\[ a^p=b!+p. \]](http://latex.artofproblemsolving.com/9/f/6/9f674dabcb364872cef4153198e2e362ae40a2fd.png)
the set of all positive integers. Find all functions 
such that for all positive integers 
and , the integer 
is nonzero and divides 
.
, there are exactly two or three positive real numbers 
satisfying the equation 
.
, 
. Points 
lie on segments 
respectively. Points 
are such that 
and 
are equlateral, 
and 
lie on opposite sides of 
, 
and 
lie on the opposite siedes of 
. Prove that
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
with 
. Prove that:![\[
\sqrt{a + (b - c)^2} + \sqrt{b + (c - a)^2} + \sqrt{c + (a - b)^2} \geq \sqrt{3},
\]](http://latex.artofproblemsolving.com/c/b/5/cb551e6d6ce94902b7bcb78f6da210d6c27b96d5.png)
with equality if and only if 
.
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.
![$P(x) \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/4/5/3/453a624c3b002c1b0e78e0023b24dd22ddd03557.png)
, with degree 
, such that 
for all 
. If the sum of all roots of 
is equal to 
with 
coprime, then what is the product 
?
(with 
)
in 
is 
and must be zero and so 
now note that 
is a zero poly.
be 
so the roots of 
are also 
s .Summing we get 
.
Where 
is a root of 
then so is 
.
is odd, this forces one root of 
and the others can be paired up into roots summing to one. Thus sum of roots is 
and answer is 
.
we see that coefficient of 
which basically gives us 