AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
+1 w
where 
is a positive integer and 
is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
such that 
divides 
for all positive integers 
and 
with 
.
, are there infinitely many positive integers 
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
be an acute triangle with 
, Let 
be the intouch triangle with 
,
,
,, let 
be the intersecttion of the perpendicular from 
to 
with 
, and 
.
and 
are concylic
the set 
can be partitioned into triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle. be a triangle with circumcircle 
, circumcenter 
, and orthocenter 
. Assume that 
and that 
. Let 
and 
be the midpoints of sides and 
, respectively, and let 
and 
be the feet of the altitudes from 
and 
in 
, respectively. Let 
be the intersection of line 
with the tangent line to at 
. Let 
be the intersection point, other than , of with the circumcircle of 
. Let 
be the intersection of lines 
and . Prove that 
.
be the foot of perpendicular from to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle . A circle 
with centre 
passes through and , and it intersects sides and at and 
respectively. Let be the foot of altitude from to 
, and let be the midpoint of . Prove that the circumcentre of triangle 
is equidistant from and .
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
, prove that is a square.
, 
, 
, 
satisfy 
and 
. Prove that there exists a permutation 
of 
such that 
.
be a cyclic quadrilateral. A circle centered at passes through and and meets lines 
and again at points and (distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
are similar.
such that 
, where 
are non-negative reals such that 
.
.
and 
then 
so WLOG 
is multiple of 3 so lets prove by induction that 
is multiple of 3 is and we know that the coefficient of 
in 
is 0 so 
or ![$b_{0}a_{1}=-b_{1}a_{0]}$](http://latex.artofproblemsolving.com/3/d/c/3dc4a0740350f19c79c121fc8657824032ba44e0.png)
and 
is not multiple of 3 so 
must be and that way you can get the result.
in and having that 
for 
all that is sufficient if the coefficient is 0 in 
and doesnt occur if 
if that is true then 
with g not multiple of 3. so then 
for some integrer g, but then 
divides 3, so 
or 
and both are wrong 
?
is irreducible over 
. Since largest degree's term is 
, its factors' largest degree's terms are . By the Extended Eisenstein Criterion, has an irreducible factor over with degree larger that 
. So it has an irreducible factor over with degree 
or 
. Let's suppose has an irreducible factor with degree . So it also has an irreducible factor with degree , and since its largest degree's term is , it has a zero over , which means also has a zero over . But is always odd, therefore it can't be 
. Contradiction! So has a irreducible factor over with degree , which is , QED.
(which is prime) and 
(which is squarefree and not divisible by ). And consider the case 
separately.
(mod 3). Simplifying gives 
. Suppose f(x)=g(x)h(x), for some ![$g, h \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/e/c/2/ec21fa9fb14a6bb2c3ab471505d593c26ae896e6.png)
nonconstant. Since x and x-2 are irreducible in ![$\mathbb{F}_3[x]$](http://latex.artofproblemsolving.com/5/e/e/5eef3e54fa50c9efdeaff24f3eb659eee7b1567a.png)
, we have (WLOG) 
and 
, for some integer polynomials 
and 
. First we consider the case where k=0. This means that g is nonconstant, a contradiction. If k=n-1, then h(x) is a linear polynomial, so f must have an integer root. We can easily check that this is false. Finally, if 1<k<n-1, multiplying g and h gives 
. Setting this equal to our original polynomial and plugging in x=0 gives 
. Since ![$g_1, h_1 \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/6/8/8/688cdb82d53e21113042d8df065330c246066cc8.png)
, this is clearly a contradiction. Thus, our polynomial is irreducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
, you deduce 
, which is correct, only that you write 
. Nice piece of algebra! All we have is 
. And now we will have 
, i.e. 
; obviously, since 
are the factors of 
.
is clearly wrong, since at least one of them must be 
. Moreover, in your mistaken way, you miss the possibility 
.
, which you then write 
. True, 
. But you don't know that the 
's, 
are distinct, and even if they were, this does not contradict that Fundamental Theorem (rather the simple theorem that a polynomial cannot have more roots than its degree), since 
is not just one value; for example for the polynomial 
we have 
and 
. So even if your argumentation up to this point were correct, you still have no contradiction. When only knowing , you need 
for at least 
distinct arguments, in order to get a contradiction.
. We show that all the a's are divisible by 3 and using that we will establish a contradiction .[Here , we will show using only + ve sign of 3 , the other case is
, in which constant terms contains only - ve terms and the casee follows only the + ve case . So , we will show here only the first case] .
, which is false since 3+5 = 8.
, which is false since 3 + 5 = 8.
n - 2, and hence the coefficients of x, 
are all zero. Since the coefficient of x is zero, we have: 
, so 
is divisible by 3.
are all divisible by 3. Then
. If s-1 
t+1, then 
= linear combination of 
. If s-1 t+1, then = linear combination of some or all of 
is divisible by 3.
,
which gives us that all the a's are multiples of 3. Now consider the coefficientof 
which is also zero.
be an integer and let 
Prove that there do not exist polynomials 
each having integer coefficients and degree at least one, such that 
is a greater number than 
. Hence, we are done since Perron's finishes over ![$\mathbb{Z}\left[x\right]$](http://latex.artofproblemsolving.com/f/5/2/f52cd80f01fabfbb6a7c7f79c0b3b5ad1762a232.png)
. This means that has only 1 root of modulus greater than 1, which means either 
or 
has roots not outside the unit circle. You'll reach a contradiction that 
yet 
is, by definition, a nonnegative integer.
be a polynomial with integer coefficients such that 
and ![\[|a_{n-1}|>1+|a_{n-2}+\cdots+|a_1|+|a_0|.\]](http://latex.artofproblemsolving.com/b/f/6/bf6a55cb8ced7afc5562c3f09a5e35f23224e90d.png)
Then 
is irreducible in ![$\mathbb{Z}[X].$](http://latex.artofproblemsolving.com/4/3/1/43177351d8a84b6c3bbfa4d03b61aa98735ac713.png)
is irreducible in 
so the polynomial is irreducible as required. Below is an alternate solution through Eisenstein's Criterion, of which here is the statement:
be a polynomial with integer coefficients. Then is irreducible in ![$\mathbb{Z}[X]$](http://latex.artofproblemsolving.com/3/c/3/3c382e829ea7c95d18cfb978f22c8c5429d5c767.png)
if there exists a prime such that
and
Assume on the contrary that 
Taking mod three, we get ![\[f(X)g(X)\equiv X^{n-1}(X-1)\pmod 3.\]](http://latex.artofproblemsolving.com/4/7/c/47cb6b39f9edc1f53fafe7eae53381f41d63254e.png)
Let 
Now, 
since 
both have degree at least two. Thus, we have 
and 
so 
which is clearly false because 
for some polynomials ![$g\in \mathbb Z[x]$](http://latex.artofproblemsolving.com/0/5/6/056648d74d347c5435c20b3d299a70de7e959500.png)
and ![$h\in \mathbb Z[x]$](http://latex.artofproblemsolving.com/f/2/4/f242450ef9b58da613f08db111320d9f1caf1fb8.png)
. Plugging in 
yields 
, so without loss of generality 
. However, taking modulo 
yields![\[
x^n - x^{n-1} \equiv \bar g(x) \bar h(x)\pmod 3.
\]](http://latex.artofproblemsolving.com/8/3/7/8372df537bbe4c89214e6bff812de16ac8538cf0.png)
Thus, 
cannot divide 
.
and 
for some polynomials ![$p\in \mathbb Z[x]$](http://latex.artofproblemsolving.com/e/6/0/e6052be3f791e17ef69d36689147ea9a274a9468.png)
and ![$q\in\mathbb Z[x]$](http://latex.artofproblemsolving.com/c/a/f/caf2037a52f444631f2e200426d58cb1bfe68910.png)
. Since 
and 
, these inequalities are actually equalities. That is, is linear and has a rational root. But the only rational roots can possibly have are 
and 
(RRT combined with the fact that has no positive roots), and both of these fail.
and 
for some polynomials and . Comparing degrees in the same way implies that is a constant. In turn, is irreducible.
where ![$h(x),g(x) \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/e/4/3/e430471ebc51ee66505fd508fd47cda3ca8d16d3.png)
. plugging in 
yields 
and plugging 
we get 
.
hence 
and 
are not linear.
where 
and 
's are root of for 
( since ![$h(x) \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/5/e/1/5e1261f96cc148d56d09e638155055eb4f940bfd.png)
)
we get 
hence we get:
which gives that 
( as and ![$h(x)\in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/b/6/2/b62f78bf3e171b8df8f9a1ab2e2d6898f093497e.png)
) we have that 
for 
which gives a contradiction , and hence the result follows 
is irreducible in 
since it is primitive. Observe that 
is irreducible in by Eisenstein. Then ![$x^n+p \in \mathbb{Q}[x][x]$](http://latex.artofproblemsolving.com/1/e/4/1e46f78e26661133aadab9649e570572c9e1e146.png)
, where is taken as the constant coefficient, is irreducible in ![$\mathbb{Q}[x]$](http://latex.artofproblemsolving.com/3/d/9/3d963b0224d7c8b822ad2df9d699fd043f0013e8.png)
by the general Eisenstein criterion, hence is irreducible in .
is any irreducible polynomial of degree less than 
, then 
is also irreducible. This is clearly false, take 
and 
.![$\mathbb Q[x][x]$](http://latex.artofproblemsolving.com/4/1/9/4199d5c9854a1d22a769de61b85fe39c557ecc9b.png)
? If you mean "the ring of polynomials in whose coefficients are polynomials in ", then this really is the same thing as ![$\mathbb Q[x]$](http://latex.artofproblemsolving.com/6/1/4/614d3ebbd60142eaf2b4d95f9bb08e7f957ffb96.png)
. In particular, given a ring 
, the polynomial ring ![$K[X]$](http://latex.artofproblemsolving.com/b/8/f/b8f4ddba89151fa31ddf44f6c22015162f2b58b0.png)
can be thought of taking the ring and adding a new variable which commutes with elements of but otherwise has no additional properties. This means something like is inherently problematic, because the first and the second interact with each other nontrivially.)
such that 
.
.
.
.
be the roots of 
.
for all 
,![\[\vert g(-5)\vert = \vert (r_1+5)(r_2+5)\dots (r_m+5)\vert = 3^m.\]](http://latex.artofproblemsolving.com/3/c/c/3ccdb265a0d6fd0d814aeffa88097d864092ba2f.png)
But 
,
RRT 
.
,
,
.
.
,
,
,
.
be the degree of 
,
,
.
in 
.
.
,
,
is irreducible, so 
reducible, so by contrapositive, 
,
and had 
,
being irreducible doesn't imply 
is irreducible, since 
)
![$\mathbb Z[x]$](http://latex.artofproblemsolving.com/b/d/c/bdc78abe277078318648d69381180248b72777be.png)
.
to divide every coefficient except the leading one. Because