[/quote]
,
be a 
-
.
,
in 
such that the sets ![\[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \]](http://latex.artofproblemsolving.com/d/5/c/d5c2a7b943e8c7200361698f3eb6f83700ad68d6.png)
are pairwise disjoint.
be a triangle. Let 
be the feet of the altitudes from 
respectively. Let 
be the projections of 
.
.
and 
Prove that
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
,
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
such that
such that
Prove that
such that
such that for any set of consecutive integers taken from the given set of integers, there will always be a colour colouring exactly one of them? That is, for all integers 
such that 
,
.
is given. It turns out that for every positive integer number 
,
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
is self-bounded if for each 
,
there exist at least 
terms of the sequence that are less than or equal to 
for a self-bounded sequence 
maps the zero polynomial to itself,![$P \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/f/7/c/f7c4c970908aa5570a9afcc45fc16c1d36cc3d5d.png)
,
,![$P, Q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
,
and 
have the same set of real roots.
Y by GorgonMathDota, Superguy, Rg230403, Aritra12, Euler1728, p_square, starchan, Hexagon_6-, biomathematics, sqrtX, Abidabi, fidgetboss_4000, CaptainLevi16, ubermensch, Williamgolly, Rounak_iitr, TSM_Zeki_MuREN
Let ![$\mathbb{R}[x]$](//latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
be the set of all polynomials with real coefficients. Find all functions ![$f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$](//latex.artofproblemsolving.com/0/3/3/033be3ab043bd47e537f01791c813739ad5c7da5.png)
satisfying the following conditions:
Proposed by Anant Mudgal, Sutanay Bhattacharya, Pulkit Sinha
![$\mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
be the set of all polynomials with real coefficients. Find all functions ![$f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$](http://latex.artofproblemsolving.com/0/3/3/033be3ab043bd47e537f01791c813739ad5c7da5.png)
satisfying the following conditions:
maps the zero polynomial to itself,- for any non-zero polynomial
![$P \in \mathbb{R}[x]$](//latex.artofproblemsolving.com/f/7/c/f7c4c970908aa5570a9afcc45fc16c1d36cc3d5d.png)
, 
, and - for any two polynomials
![$P, Q \in \mathbb{R}[x]$](//latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
, the polynomials 
and 
have the same set of real roots.
Proposed by Anant Mudgal, Sutanay Bhattacharya, Pulkit Sinha
This post has been edited 1 time. Last edited by anantmudgal09, Mar 7, 2021, 5:20 PM
Reason: I had to do this after Ankoganit suggested it. Missed opportunity.
Reason: I had to do this after Ankoganit suggested it. Missed opportunity.
?
,
between that of 
,
for certain types of Q (aka ones with distinct roots), hence forcing 
for all 
having equal roots, forcing 
to be fixed for all polynomials and finally forcing 
should work?
and 
are two valid functions. We will show that no other functions work. Pick a function 
in (iii), we see that 
and 
have the same set of real roots, hence 
.
on 
if 
and 
have the same set of real roots. Thus, 
and 
implies 
for all ![$p \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/5/0/f/50f6547993a8cd5fdab4a0c85439f848c5929dcf.png)
.
,
for all ![$p, q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/0/0/a/00ab98ffa310672c06575a07a2156b675b64870f.png)
.
by (ii), so we see that if 
then 
.
for all 
,
is even. Choosing a constant 
such that 
has no real roots, then 
hence 
cannot have an odd degree, so 
for all 
,
always has a real root. Plug 
for 
,
always has a real root for all 
;
is odd, hence equal to 
.
.
,
implies that for some real number 
,
.![$q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/3/6/2/362c6120ddd5f03d4bd178b8a688cc8c052b13f4.png)
,
with degree bigger than 
are chuunibyou.![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
.
and is bigger than 
.
,
,
be the extrema of ![$[c_1,c_2],\cdots [c_{n-1},c_n]$](http://latex.artofproblemsolving.com/b/5/5/b5522bcf008bb9dcf2511dfe94646ce5ce4c35b8.png)
in that order, and suppose they are attained at 
Make 
large enough so that ![$|k_i|> \max_{x\in [a,b]}|p(x)|$](http://latex.artofproblemsolving.com/e/f/5/ef52b331af23f1dc1587d830777648b2c67e0de9.png)
for all 
has degree 
for 
,
.
![$r \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/6/c/5/6c5253a9acd0d8387ea0aff74c6e20fbdc6dc87a.png)
,
so 
for some constant 
.
hence 
,
is a constant such that 
for any chuunibyou 
for all 
.
if 
has a real root which is not a root of 
Indeed, if 
is a root of 
)
,
.
such that 
.
is such that 
and 
both have real roots, so 
.
,
,
hence 
,
or 
in (3) to get that 
and 
and 
in (3) .
then 
for some non zero real 
.
)
and 
, let 
(by claim 4 it is constant polynomial)
and 
are friends , but latter can only have 
real roots and former has exactly one , hence contradiction.
.
is a constant polynomial 
and 
, so 
and 
are friends ; latter has root 
and so former must also have same root.
and hence 
and 
is already dealt with.
or less.
of degree 
be 
, let 
with sum equal to 
, so that 
has degree at most 
.
and 
are friends.
and hence same logic gives that degree of 
and let 
be distinct reals .
so that 
has degree at most 
and 
are friends , but 
we have 
and so 
and so the only possibility is 
'
is not entirely necessary. I did use this in my official write-up but not in a crucial way. Even the condition 
can probably be relaxed, if one includes the solutions 
in the solution set. I'll try to finish that later, for now a solution that assumes everything the problem gives.
and 
,
are sweet.
are bounded by 
on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
,
by 
for 
using Lagrange Interpolation. Now, each of the polynomials 
by intermediate value theorem, and hence is sweet.
and by matching the leading two coefficients, 
and 
.
is a constant. It is not hard to analyze to find out that this constant must be 
.
and 
have the same set of real roots. But now this set of real roots must also be a root of 
but as we can vary 
and generate arbitrary roots of 
.
,
have same set of real roots and thus these are also roots of 
and again by same logic, we have 
.
Let 
If 
)
we have 
has the same real roots as 
.
is at most quadratic which has the same unique real root as 
so of the form 
or 
with 
we have 
is one of 
Note that 
had the same degree as 
Then for a nonzero constant 
of the opposite sign of 
had no real roots but 
had. Note that, 
for each linear polynomial 
.
has the same root as 
.
.
of degree 1 at most. Note that 
either 
for these polynomials.
be a polynomial degree 
and suppose ![$P\mathbb{R}[x]^{\le n-2} $](http://latex.artofproblemsolving.com/8/e/7/8e7180bd5df16e04f6613162193cebe35487a841.png)
of its translated by the polynomials of degree at most 
real roots then either has repeated real roots or an irreducible quadratic factor . We have 
where 
has degree 
and 
is quadratic with at most one real root. Choose 
such that 
has 
has more real roots than P, contradiction.
has 
be one of 
and by induction forall the polynomials 
we have 
Suppose 
has degree 
By the lemma there is 
has 
requires 
correspond to condition 3.
is bijective.
have the same set of real roots. Plugging in 
have the same set of real roots.
.
,
is even, we can choose 
has no real roots, whereas 
Contradiction.
for 
.
for distinct 
.
have the same set of real roots 
.
have the same set of real roots (say 
)
.
for some variable real 
with pairwise distinct roots, as above. Consider 
must have the same set of real roots. Similarly, 
must have the same set of real roots. As we can let 
to be as large as we want along with the fact that 
is a root of 
and 
,
,
.
.
and 
and the third assertion by 
.
have the same set of real roots. So 
and so 
have the same real roots (
.
.
(Here 
is any constant polynomial).
can’t be 
,
and 
to not have the same set of real root for some 
have the different sign from 
for some 
.
.
we have that 
and 
which is a clear contradiction. 
have the same set of real roots. So 
which means that 
which is a clear contradiction.
have different roots .Consider 
which gives that 
have the same set of roots, which means that there’s a constant 
such that 
and since 
and 
.
.
with different root from both of them and let 
, then we have 
and so 
.
which is 
.
.
but 
.
gives that 
and 
have the same set of real roots which contradicts their degree being equal to 
![$\forall R \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/b/7/ab7727b9a0610bb91ba16a7e7d07ef70d1e687a0.png)
or 
have the same set of real roots. So now varying ![$$\boxed{f(Q) = Q \text{ } \forall Q \in \mathbb{R}[x]}$$](http://latex.artofproblemsolving.com/0/5/4/05449b23eba6b7ef501981f08d52ca2d778a60dd.png)
which indeed is a solution.
have the same set of real roots. So now varying ![$$\boxed{f(Q) = -Q$ $\forall Q \in \mathbb{R}[x]}$$](http://latex.artofproblemsolving.com/6/9/3/693dd96a4cb3d95c5ad0e4ad580f18d6f3d6b1b9.png)
which also is a solution.
is constant, 
is also constant and 
.
shows that 
iff 
,
for all 
.
this shows that 
and hence 
is bijective.
(since 
and hence 
.
gives 
gives 
.
,
for a good polynomial 
,
and 
.
.
is the same for all 
depending on the parity of the degree of 
into 
due to involutivity. Now use good polynomials 
here to get 
.![$Q\in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/8/a/a8ad590d9f5fe2fefa7422ee234e24ee3f844cc7.png)
,
with 
.
.
,
)
for 
.
has no real root, note that any degree 1 polynomial has a real root, hence 
for some 
.
refers to the polynomial 
and the degree condition, 
can only be 
or 
.
,
and 
gives (1) 
is degree 1, which has 1 root; and (2) 
has two root 
for any 
gives 
has the same root as 
.
for any polynomial 
.
,
.
has the same real real root set as 
.
is a solution of 
,
.
.
,
.
![$f~:~R[X] \rightarrow R[X]$](http://latex.artofproblemsolving.com/b/1/5/b15ecba0a1d0e0c14ff512bb6df2127509f6155d.png)
, which satisfies :
,![$P~\in~R[X]~,~degf(P)~\le~1+degP$](http://latex.artofproblemsolving.com/9/b/d/9bd8434fe3208f03f9cc337e733c281a24df02fe.png)
, and![$P,Q \in R[X]$](http://latex.artofproblemsolving.com/6/5/c/65c43fc3c032d1421bf65b24a125437ef81a25f4.png)
, the polynomials 
have the same set of real zeroes.
, we get that , 
and 
have the same set of real zeroes.Hence , 
, (otherwise we get a contradiction on the maximality of zeroes of 
, where 
"
"
and 
are only such functions which satisfy the above conditions .
, hence we are left with with symmetric inequalities : 
.
, thus 
.Lets consider 
, then its a linear polynomial which has exactly one real zero and hence we see that 
doesn't have same set of real roots , which is a contradiction , hence 
.
, then if 
is even , then obviously 
has to be odd , and 
-
, now consider 
which is a constant polynomial where 
are the points of local minima's of 
has no real zeroes , but 
is of odd degree and should have at least one real zero , hence a contradiction.A similar type of argument follows for the case where 
.
if P and Q have the same real roots.
.
.
,
.
,
or 
,
.
have the same signal, but you can put 
or 
.
,
is constant, so, 
and 
.
(just put 
)
.
,
for an infinite number of 
.
,
denote the assertion of the second bullet point, and 
of the third. From 
,
is an involution. Now, 
tells us that 
and 
have the same set of real roots, implying that 
,
is a real constant, but 
tells us that the degree of 
is at most 
,
or 
.
,
is a constant gives us that 
and 
have the same set of real roots, and since 
is either constant or linear by 
,
for all constants 
and 
and 
have the same real roots. If 
to get infinitely many values for ![\[C=f(f(C))=\frac{C}{c^2}\]](http://latex.artofproblemsolving.com/b/e/5/be59bdb42bd76d968bb09dbd261dcb0582dd21d0.png)
implying that 
.
,
and 
,
gives us that 
and 
have the same set of real roots, implying that 
have the same set of real roots. Fixing 
for all polynomials ![\(P\in\mathbb{R}[x]\)](http://latex.artofproblemsolving.com/2/e/9/2e9d719a629c13bae85d6cfa58875a34a9b2f1ea.png)
.
,
for all polynomials ![\[f(P)=P~\text{or}~f(P)=-P\forall P\in\mathbb{R}[x]\]](http://latex.artofproblemsolving.com/4/a/6/4a6bbbed88df12dd64615e71342b003568331241.png)
.
is even, there exists some choice of constant 
should have no real roots, i.e. its degree should be even too, hence 
which is odd. Then due to the fact that 
is odd then 
is even, so 
.
for some 
.
,
has all roots real and distinct as well, and 
has the same roots. Since their degrees are the same, it follows that they are multiples of each other, i.e. 
.
and 
to get that 
,
and 
we may choose some 
whose roots are all real and distinct. Then for sufficiently large real 
,
are also all real and distinct, and the same as the roots of 
.
then the polynomials 
have the same leading coefficient, degree, and set of 
roots, hence they are the same, so 
then the polynomials 
have opposite leading coefficient, and the same degree and set of 
in the third assertion. This tells us that 
.
for every constant polynomial 
in the third assertion. This says that 
.
(otherwise swap 
and use 
)
.
in the third assertion. This means that 
and 
have the same set of real roots. Clearly 
is not a constant or linear, because then 
or 
.
in the third assertion. This means that 
and 
have the same set of real roots. The real roots of 
,
,
for any constant 
.
.
was cubic, an almost identical analysis as case 1 gives a contradiction. So suppose that 
gives that 
,
gives that 
.
.
and noting that 
gives that 
.
for two distinct non-zero constants 
.
first, giving 
and then let 
giving 
.
.
.
,
giving 
and then put non-zero 
for non-zero 
and 
for distinct non-zero 
in the third assertion, we get that 
have the same set of real roots for \textit{any} real number 
,
has 
,
for every 
.
for every constant 
and ![$P\in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/7/4/c/74cb2247069abb8a22666e87fcbd69a55a015004.png)
denote 
.
and 
.
and 
,
.
,
so 
,
for constant polynomial 
to get 
,
.
is a polynomial, 
.
,
where 
.
,
,
and 
where 
