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.
be the lengths of the sides of a triangle. Prove that![\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]](http://latex.artofproblemsolving.com/7/4/b/74b7aa5621bc52977d9acd448d74293dc27a8e00.png)
and 
be arbitrary points on the sides 
and 
of triangle 
, respectively. The circumcircle of 
meets for the second time the circumcircle of 
at point 
. Line 
meets line 
at point 
, while line 
meets line 
at point 
. Prove that lines 
and 
pass through the same point.
with circumcenter 
. 
and 
intersect the altitude from 
to at points 
and 
respectively. 
is the circumcenter of triangle 
and 
is the reflection of over . 
is the second intersection of circumcircles of triangles 
and 
. Show that 
are collinear.
and 
Find the value of 
and 
such that 
and 
is a prime number
such that 
for all 
be an even integer and let 
be real numbers satisfying 
.
be a positive integer. Ana and Banana play a game. Banana thinks of a function 
and a prime number 
. He tells Ana that 
is nonconstant, 
, and 
for all integers 
. Ana's goal is to determine the value of . She writes down integers 
. After seeing this list, Banana writes down 
in order. Ana wins if she can determine the value of from this information. Find the smallest value of for which Ana has a winning strategy.![$f\in C^1[a,b]$](http://latex.artofproblemsolving.com/8/4/8/848da12534d564560fe1cb016142159234be98b4.png)
, 
and suppose that 
, 
, is such that
for all ![$x\in [a,b]$](http://latex.artofproblemsolving.com/7/8/d/78d9dcd969a2c45f7cc9234aa1377b6ec1c2d5d3.png)
. Is it true that 
for all ?
![$f \in C([0,1];[0,1])$](http://latex.artofproblemsolving.com/4/7/5/475518b7db0b787dd90b93f602c7b912c663fba3.png)
bijective, 
and ![$(a_k) \in [0,1]^\mathbb N$](http://latex.artofproblemsolving.com/a/d/8/ad8a80d847c60ee8e916a1f9b08b7b03ec17c814.png)
with 
converge.
converge?
is called perfect if the sum of its positive divisors 
is twice , that is, 
. For example, 
is a perfect number since the sum of its positive divisors is 
, which is twice . Prove that if is a positive perfect integer, then:![\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]](http://latex.artofproblemsolving.com/0/c/0/0c08f527e87f7aae4ca5226f19bd084249551fe3.png)
where the sums are taken over all prime divisors 
of .
where , 
, and 
are nonnegative integers.
. Note that 
, 
, 
, and 
. Thus 
there exist nonnegative integers ,, and such that 
. Note that 
implies 
but then if 
is congruent to 
modulo 
it must be congruent to modulo 
as well as we inspect on the 
cases where the set of residues for , , and is 
,
,
, or 
.
with 
. Let 
. These constructions suffice:
Clearly 
is a nonnegative integer (AM-GM, 
is a ring). So it suffices to show that 
is impossible.
. If 
then 
, so 
. So 
. Then 
is divisible by , so 
. But then .
. We claim 
can take on any nonnegative value that isn't 
or 
. Note that plugging in 
gives 
, plugging in 
gives 
, plugging in 
gives 
, and plugging in 
gives . is divisible by , then it is also divisible by . This is true because we can factor as![\[T=(A+B+C)((A+B+C)^2-3(AB+BC+CA)).\]](http://latex.artofproblemsolving.com/e/a/3/ea375bca4fa257115092cf95765c5600547fcf5a.png)
It is evident that if divides one of the factors, it must divide the other, so 
, as desired. This completes the solution.
![$f(x) \in \mathbb{Q}[x]$](http://latex.artofproblemsolving.com/e/d/a/eda31c06f8b59911d0cedd1128bbd5499094fc2c.png)
and 
is an element of ![$R := \mathbb{Q}[x]/(f(x))$](http://latex.artofproblemsolving.com/4/9/d/49d336ecc1bf5fe0cc84202407461abf69c1b5ff.png)
, then the Norm of is: taken over all 
roots of 
-linear map 
given by multiplication by .
is the Norm of 
when we take 
. Since has two factors over and three factors over 
, the same will also be true for . The fact that ![$\mathbb{N}[x]/(x^3-1)$](http://latex.artofproblemsolving.com/e/f/4/ef492e47d0fdfd9a66b9ddf7b160a90d395d432d.png)
is closed under multiplication implies that its set of Norms (the set of values in question A1) is also closed under multiplication. This may be a good way to construct similar exercises.
![\[a^3+b^3+c^3-3abc=\frac 12\cdot (a+b+c)\sum_{\mbox{cyc}}(a-b)^2.\]](http://latex.artofproblemsolving.com/9/8/c/98cdc7d9bb55c6ed37113ec2fc45ac992d8e279c.png)
So by taking 
we can achieve any 
and by taking 
we can achieve any 
. It remains to figure out which multiples of we can achieve. We can achieve multiples of by setting 
to get 
.. Then exactly one of 
and 
is a multiple of . If the latter is the case: if some two of are the same mod , then all are, but this cannot occur so are all distinct modulo . This is also a contradiction. So then the first case holds. Then are all congruent modulo or all distinct modulo which again ends up causing a contradiction. and multiples of are achievable.
and Introduction to Number Theory level Modular Arithmetic? If so, can someone provide a basic solution like that? Just wondering.
as residues 
such that everything else works?
residues. is a multiple of . We claim that 
and 
, which together imply that 
.
by FLT, so .
and permutations then . These are the only possibilities since . can only be 
. We can show that all of these are achievable:
, is possible.
, 
(except for ) is possible.
, 
is possible.
, 
is possible.
, 
is possible.
, 
is possible.
, 
is possible.
, 
is possible.
.
..
in , and then show 
.
(indeed, this covers 
mod ).
works.
(indeed, this covers 
mod ).
works.
, set 
. This works. For 
, set 
.
or 
.![\[f(A,B,C)=(A+B+C)(A^2+B^2+C^2-AB-BC-AC)=(A+B+C)((A+B+C)^2-3(AB+BC+AC))\]](http://latex.artofproblemsolving.com/c/7/d/c7d7c28b8d40b487a1ccd24d5bed182a19e99cb1.png)
must divide and 
. If divides 
, then divides , a contradiction. If 
, then , a contradiction.
.![\[f(A,B,C)=(A+B+C)(A^2+B^2+C^2-AB-BC-AC)\]](http://latex.artofproblemsolving.com/3/d/5/3d5cec918d9d55efa51b68e1fca0cde90998a0cd.png)
, because 
.
.
, which implies is nonnegative. 
that are either 
or 
mod 3 or mod 9.
, then 
is a construction
, then 
is a construction
, then 
is a construction is divisible by , it must be divisible by . This can be seen by factoring and looking over all residue combinations of 
módulo that allow the expression to be divisible by . Thus, these are all the cases.
with 
.
, we get .
, we get 
. Similarly, by taking 
, we get 
.
, we get 
. 
and 
.
to get the result. 
. while . Finally, . We now show that it is impossible for 
.
and 
is possible only if:
,
.
, then 
and 
which means 
. Hence, 
.
.
then 
. If 
then 
, while 
. Hence, . then , as claimed.
work for every 
.
work. Let's see the constructs for these values:
(constitutes for 
) for all non negative integers .
(constitutes for 
) for all positive integers .
(constitutes for 
) for all non negative integers .
Thus 
are not achievable. 
works.
First we prove if 
then 
, notice 
, now 
, now that forces 
, now that implies 
for some 
and hence 
, so clearly 
are not possible. 
gives 
under , which gives as residues under .
we get 
which gives residues under , and finally for 
we get , which finally accounts all possible values of for non negative integers . 
is a multiple of 3, denote 
must be a multiple of .
, denote 
, where the integers in the parenthesis are the remainders of divided by . Test three cases, 
, 
yielding 
as desired.
gives 
,
gives 
, and
gives 
. Letting 
vary gives all solutions.
, 
follows by inspection.
Thus, we can obtain all integers 
such that 
.
Thus, we cannot obtain numbers that are or 
, and we are done.
or 
mod . Indeed, by setting 
we get 
which covers all multiples of , and by setting 
we get 
which covers all non-multiples of . or mod integers fail. This just stems from the observation that![\[a^2+b^2+c^2-ab-bc-ca = (a+b+c)^2 - 3(ab+bc+ca) \equiv (a+b+c)^2 \pmod 3\]](http://latex.artofproblemsolving.com/8/1/0/8104b784cf4fd2f5fd7f86d0e8ce0826200ccb74.png)
so either 
is the product of two multiples of or two numbers relatively prime to .
such that 
. To construct all possible values, we have
to get 
,
to get 
,
to get 
,
to get 
.
covers all numbers that are not 
.
, we have
is impossible.
such that . We know that,![\[f(a,b,c)=a^3+b^3+c^3-3abc = \frac{1}{2}(a+b+c)\left((a-b)^2+(b-c)^2 + (c-a)^2\right)\]](http://latex.artofproblemsolving.com/e/9/e/e9e8c7655408fbcde055f2baa20fddbc1788bd34.png)
Hence, clearly for non-negative integers always 
is non-negative. Then note,![\[f(x+1,x,x) = \frac{1}{2}(3x+1)(1^2+1^2+0^2) = 3x+1\]](http://latex.artofproblemsolving.com/0/e/1/0e1627ddae56766e754a40b33c0fe7c1a2892630.png)
for all non-negative integers and![\[f(x+1,x+1,x) = \frac{1}{2}(3x+2)(1^2+1^2+0^2)= 3x+2\]](http://latex.artofproblemsolving.com/e/2/c/e2c6f48bd4a54fd7369db4929271da6c27d6ca8f.png)
for all non-negative integers . Further note that since 
if 
and 
for all 
, if we let 
, 
and 
for integers 
then![\[a^3 \equiv \epsilon_a \pmod{9}\ , \ b^3 \equiv \epsilon_b \pmod{9}\ , \ c^3 \equiv \epsilon_c \pmod{9}\]](http://latex.artofproblemsolving.com/8/4/1/84192c6c0a094e167e90cf7cbb87e00e0ddf56a4.png)
Moreover,![\[a^3+b^3+c^3-3abc \equiv a+b+c \pmod{3}\]](http://latex.artofproblemsolving.com/a/3/b/a3b1e2eb39cbb41082d6a6d903f148ecd9ef70df.png)
so 
if and only if 
. However due to our previous observation, if we have![\[a+b+c \equiv \epsilon_a + \epsilon_b + \epsilon_c =0 \pmod{9}\]](http://latex.artofproblemsolving.com/a/4/d/a4d91d9a4c74a9a2a710b823f6791cf1c61f2fae.png)
Hence if we must have 
. To finish note that,![\[f(x+2,x+1,x) = \frac{1}{2}(3x+3)(2^2+1^2+1^2) = 9(x+1)\]](http://latex.artofproblemsolving.com/c/5/d/c5d86101366df53375e24d4fea81b590ac8a1db7.png)
for all non-negative integers . Finally when 
we have 
so the answer must indeed be as claimed.
and 
are achievable, we will show that 
,
implies 
.
iff 
,
iff 
,
iff 
.
,
,
,
are 0, -1, 1 
(in some order). Using the above observation, it can be checked that cases i) and ii) imply 
.
equivalent to 
.
.
gives 
and 
gives 
,
gives 
.
work. To construct you just take 
,
,
.
.
