AoPS Academy
be a scalene triangle with incircle 
. Denote by 
the midpoint of arc 
in the circumcircle of , and by 
the point where the 
-excircle touches 
. Suppose the circumcircle of 
meets again at 
and intersects at two points 
, 
.
or 
is tangent to .
, with 
, for which the edges 
are
with 
, for which the edges 
are all coloured blue.
for which the value of the expression![\[x^2+6x+33\]](http://latex.artofproblemsolving.com/3/d/e/3de5bc2a9b022192315975731d48af7fda9fe228.png)
is a perfect square.
for which there exist infinitely many positive integers 
such that ![\[ \frac{a^m+a-1}{a^n+a^2-1} \]](http://latex.artofproblemsolving.com/2/8/a/28a68e409451bbdcaee7c1286341f6aeea9ffdad.png)
is itself an integer.
Two circles 
and 
with radii 
and 
respectively are externally tangent. The straight line 
is tangent to the circles 
and at points and 
respectively.The parallel line 
to the line is tangent to the circle and intersects at points 
and 
The line 
through intersects the line 
and the circle at points 
and 
respectively, different from and Prove that the circumcircle of triangle is tangent to the line 
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.
,
or 
,
is integer.
is a quadratic equation, and every quadratic with coefficients in 
for 
so 
or 
.
and 
then 
.
we have 
or 
