from the set of positive integers to the set of positive integers such that, for all positive integers 
and 
,![\[ a, f(b) \text{ and } f(b + f(a) - 1).\]](http://latex.artofproblemsolving.com/4/a/b/4ab97da27e05d0f9e7762f74d0d0bd2dc8b9842f.png)
be an acute-angled triangle with 
.
be the circumcircle of 
be the midpoint of the arc 
of 
.
meets 
at 
and meets 
.
at 
.
by 
.
.
meets line 
.
be a cyclic pentagon inscribed in a circle of centre 
which has angles 
.
and 
meet at a point belonging to the diameter 
.
and the segment 
intersects the circumcircle of 
in a point 
.
be the midpoint of 
is parallel to 
.
Y by hadikh, Amir Hossein, mostafataheri, Adventure10, Mango247
Consider the second degree polynomial 
with real coefficients. We know that the necessary and sufficient condition for this polynomial to have roots in real numbers is that its discriminant, 
be greater than or equal to zero. Note that the discriminant is also a polynomial with variables 
and 
. Prove that the same story is not true for polynomials of degree 
: Prove that there does not exist a variable polynomial 
such that:
The fourth degree polynomial
can be written as the product of four 
st degree polynomials if and only if 
. (All the coefficients are real numbers.)
Proposed by Sahand Seifnashri
with real coefficients. We know that the necessary and sufficient condition for this polynomial to have roots in real numbers is that its discriminant, 
be greater than or equal to zero. Note that the discriminant is also a polynomial with variables 
and 
. Prove that the same story is not true for polynomials of degree 
: Prove that there does not exist a variable polynomial 
such that:The fourth degree polynomial
can be written as the product of four 
st degree polynomials if and only if 
. (All the coefficients are real numbers.)Proposed by Sahand Seifnashri
This post has been edited 5 times. Last edited by goodar2006, Jun 28, 2012, 5:49 PM
.
iff these three conditions are satisfied:
plane with 
,
is nonnegative only in this region, so it's value must be 
on the boundry. So only for all nonpositive real numbers 
,
:
to be roots of 
, then by Vieta's relations
if 
and it becomes negative when 
are two conjugate non-real complex numbers (here we should note that 
)
where 
,Then 
becomes a polynomial because 
and 
for some 
on unit circle . Let 
varies on complex numbers with norm 
.For every such 
, 
,except when 
, i.e 
. In this two exceptions we should have 
, But according to continuity of 
can not be positive because if it become positive then there is a 
-
that it remains positive on it , which is absurd . Hence 
for all real 
vanish at every 
where 
, contradiction !
then 
so 
thus 
.
be the greatest positive integer s.t. 
.
so 
.
s.t. 
be arbitrary constant. Let 
then 
so 
.
then 
so 
.
is a root of 
and 
is a factor of 
was arbitrary constant, so 
.
is just 
,
where 
only for nonpositive real numbers 
,
that 
only for 
.
,
