AoPS Academy
positive real numbers satisfying 
,show that:![$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$](http://latex.artofproblemsolving.com/c/b/3/cb39e7f838c99ed8a209c6844f3d2810838d1dfd.png)
.
denote the set of natural numbers, and let 
, 
, be nine distinct tuples in 
. Show that there are three distinct elements in the set 
whose product is a perfect cube.
are written on the board.In each step,two random numbers 
and 
are chosen and deleted.Then,the number 
is written instead.What will be the number remained on the board after the last step.
in the Cartesian Plane as follows:
for some real number 
. Denote the common area of 
and 
for some be 
. Compute the algebraic form of the function for 
.
be a polynomial with integer coefficients such that 
and 
for some integer . Prove that there exists no integer such that 
.
, Where 
has a maximum value of and a minimum value of 
in the interval ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
. Confirm whether the value of 
depends on or .
, points 
, 
, and 
lie on sides 
, 
, and 
, respectively, such that
, 
, 
, 
, 
, 
.
?
) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
be a root of 
. If splits in ![$\mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/e/5/a/e5a21492095c9116b100e2f7626a70cc26834e82.png)
, then 
; otherwise 
. Either way, .
. We obtain the 
equality chain![\[\left(t + \frac{1}{t}\right)^3 - 3\left(t + \frac{1}{t}\right) + 1 = 0 \implies t^3 + \frac{1}{t^3} + 1 = 0 \implies t^9 = 1.\]](http://latex.artofproblemsolving.com/a/e/9/ae935843dc527f9cb9a88fca74b5a4f2cf4b7f12.png)
Thus the (multiplicative) order of divides 9, so it is one of 1, 3, or 9. If it is one of 1 or 3, then 
, meaning 
, so 
. Otherwise, the order of is 9. belongs to 
, a group with 
elements. (It is also cyclic, but we don't need that.) Then by Lagrange, 
, so 
. (If we know that the group is cyclic, we can directly deduce .) To conclude, either or .
trick has become quite standard [one easy application is to determine 
], and (2) we essentially need to deal with the field of 
elements, a concept not normally included in the olympiad canon.
are co-prime integers, and 
is a prime divisor of 
then 
.
and two integers such that 
, , is not a common divisor of but 
. Take the smallest such . Obviously 
, so 
. and by smaller numbers: we show that there are some integers 
, not both zero, such that 
and 
.
and 
by their greatest common divisor. Let 
, 
and 
. notice that 
and 
, so 
is not divisible by . Then the have 
.
and 
modulo 
and 
, we can see that 
or 
. Hence there is another prime divisor 
of 
with 
.
. Then
If 
then 
, contradiction.
must be checked manually.
be a root of . If splits in , then ; otherwise . Either way, .
, it has no root either in 
or 
, otherwise 
which is obviously false. Then, there is no such that 
.
be a root of . If splits in , then ; otherwise . Either way, ., it has no root either in or , otherwise which is obviously false. Then, there is no such that .
modulo .
does not represent the "field" of integers modulo 
(which can't be a field!); instead, it is the unique (up to isomorphism) degree-two extension of 
. has no root in , this polynomial is irreducible, and so the quotient ![$\tfrac{\mathbb F_5[x]}{(x^2 - 4x + 1)}$](http://latex.artofproblemsolving.com/4/b/5/4b593657c53a83056fec776871e1e8baa5778c8b.png)
is a field. We can describe this field symbolically as follows: if 
(formally!) satisfies 
, then the elements of this field are the elements of the form 
, where 
. elements up to isomorphism; we denote this field by . This construction generalizes to finite fields of arbitrary prime power.
for some 
.
has order .
This implies 
, so has order dividing . However, 
since 
. Therefore, has order exactly . 
so we're done.
for some . has order .This implies , so has order dividing . However, since . Therefore, has order exactly . so we're done.
is defined as follows: to be a root of an irreducible quadratic (say 
) in . Then is the set
with 
, which means that we can assume is the root of 
for some 
, i.e. 
for some quadratic non-residue 
mod .
in . If its discriminant 
is a quadratic residue mod , then the quadratic formula gives us two roots in 
. Otherwise, 
is a quadratic residue mod , say 
. Then notice that
.
since we divide by 
a couple of times, but we can just check it manually then.
for some . has order .This implies , so has order dividing . However, since . Therefore, has order exactly . so we're done.?
is (related to the field in question) implies in divides the order of the field. Is that what you want?
to 
. By the quadratic formula, if 
is a QR then in fact 
, otherwise is an NQR and we still have . Thus we may substitute 
, whence
where all equalities are in . This implies that the order of in is , since 
. This then implies , so we're done. 
, so that the equation becomes 
This requires the order of mod to be precisely equal to . On the other hand, 
is an element of 
, so and we have the result.
has one root in 
if , three roots in if 
, and no roots in otherwise. When , the only root is 
, from now on, assume . Throughout, let 
be a root of 
and 
be the Frobenius automorphism sending to 
. Solving the cubic shows has roots 
, 
, and 
.
, the degree of the field extension 
is 
if 
, if 
, and or 
otherwise.
if 
, which is enough since 
is Galois. We have![\[3 \ge [\mathbb F_p(\omega) : \mathbb F_p] = [\mathbb F_p(\omega) : \mathbb F_p(\omega + \omega^{-1})][\mathbb F_p(\omega + \omega^{-1}) : \mathbb F_p] \ge 2[\mathbb F_p(\omega + \omega^{-1}) : \mathbb F_p],\]](http://latex.artofproblemsolving.com/c/c/a/cca28ef38bcb80aa99a43e4a873d1490c5f366b9.png)
so 
is nontrivial and .
if , which is enough since is Galois. If , 
and we are done. If , the minimal polynomial of is![\[(x - \omega)(x - \sigma(\omega)) = x^2 - (\omega + \omega^p)x + \omega^{p + 1} = x^2 - (\omega + \omega^{-1})x + 1.\]](http://latex.artofproblemsolving.com/a/e/1/ae18d5065aea4203367a95e87869fff68fa9ceec.png)
But the minimal polynomial has coefficients in , so 
is in .
in ![$\mathbb{F}_{p^2}=\mathbb{F}_{p}[\sqrt{a^2-4}]$](http://latex.artofproblemsolving.com/1/7/a/17a67895dc1552660c66270d5cc0faa583a128d1.png)
(which has order 
)and so we have (since ) ![\[x^6+x^3+1 \equiv 0 \pmod p \iff x^9 \equiv 1 \pmod p \iff 9 \mid p^2-1\]](http://latex.artofproblemsolving.com/c/c/b/ccba0a6fb10b7c08d92175fe02972c5896d5ec94.png)
And done.
such that 
(we can't do this in modulo since 
could be a qnr). We get 
Therefore, is the order of in 
Since the order of is exactly 
it follows 
which implies the statement.
Prove that
Let 
Prove that
,
or 
,
is integer.
is a quadratic equation, and every quadratic with coefficients in 
for 
so 
or 
.
and 
then 
.
we have 
or 
