AoPS Academy
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
, 
is an homoemorphisme too.
, let 
denote the greatest prime factor of 
. A strange pair is a pair of distinct primes 
and 
such that there is no integer for which 
.
+1 w
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