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.
what is range of values of 
what is range of values of
? (there is an easy workaround)
be a sequence of nonnegative real numbers whose sum is 
. Find the maximum possible value of ![$$\sum_{k=1}^\infty \left(k\cdot (3k+2) \cdot 2^{-k} \cdot \sqrt[k]{a_1a_2\dots a_k}\right).$$](http://latex.artofproblemsolving.com/2/d/f/2df5179ce532077733530d5e9f1266c22d1cdb7d.png)
is 
, maybe even with rather elementary means? As usual, 
denotes the 
-th prime number. The problem of that limit came up in my partial solution of this problem: https://artofproblemsolving.com/community/c7h3495516.
be an integer. The sequence 
is defined via 
, 
and![\[ a_{n+2} = ka_{n+1} + a_n \]](http://latex.artofproblemsolving.com/b/5/e/b5e3e6f9f0b0709113498266e41b372299a2986f.png)
for any positive integer . Prove that there are finitely many pairs 
of positive integers such that 
.
of rational numbers such that 
.
, there exist pairwise different real numbers 
satisfying
that satisfy the equation
less than three times of their sum. Find those integers.
of positive integers with 
prime and ![\[ a^p=b!+p. \]](http://latex.artofproblemsolving.com/9/f/6/9f674dabcb364872cef4153198e2e362ae40a2fd.png)
satisfy both of the following conditions.
, if 
, then 
.
that satisfies the equation 
and 
. that satisfies the equation 
, for each integer .
has no solutions in integers 
, 
, and 
.
and such that: ![\[1^n+2^n+3^n+\cdots +n^n=x!\]](http://latex.artofproblemsolving.com/1/f/e/1fe2d44fea6376291c30257973e2e90deab76557.png)
such that , 
, 
are each the sum of the squares of two integers. [Example: 
, 
, 
.]
and 
such that 
Given such numbers we can let 
. Such numbers and render 
as a rational approximation of 
Here is a scheme for generating such numbers: Let 
and after that, 
and 
We claim that for these sequences, 
We prove this claim by induction. It is clearly true for
After that,
to be 
rather than 
we take 
and 
This gives us the following first few 
pairs: 
giving rise to these 
triples: 
This does, as required, give us an infinite list of solutions. It is very far from exhaustive, as is shown by the following list of the first 10 solution triples: 
are equivalent to 
or 
case.
let 
Clearly is the sum of two squares. Then 
so 
and 
for 
This gives us 
and so on.
and such that 
, of primitive solution 
, 
. Then we have the general solution 
, with the recurrence relation(s) 
, 
. Both sequences satisfy the linear recurrence of characteristic polynomial 
.
such that 
is the sum of two squares, and furthermore, say that there is an integer such that 
. Thus 
, so let 
for some integer , from which we find 
and ![\[ (2m^2 + 1)^2 - 1 = (2m^2)^2 + (2m)^2. \]](http://latex.artofproblemsolving.com/0/1/4/014120984f080c8428b7e7591a41b8ca560a12ff.png)
Thus one infinite family of solutions is ![\[ \{ (2m^2)^2 + (2m)^2, (2m^2 + 1)^2 + 0^2, (2m^2 + 1)^2 + 1^2 \}. \]](http://latex.artofproblemsolving.com/c/3/d/c3d904782dea46a8a1cd46d78b0208c4035df503.png)
A similar approach finds another family of solutions, ![\[ \{(2m^2 + 2m)^2 + 0^2, (2m^2 + 2m)^2 + 1^2, (2m^2 + 2m - 1)^2 + (2m+1)^2 \}. \]](http://latex.artofproblemsolving.com/7/6/3/763ae7a023d22a3878d71daca9881b7ff973d3f9.png)
There appears to be at least one more family of solutions which I haven't looked into, but the above is more than sufficient.
such that 
is a perfect square, ie 
for some . Clearly this is always possible for infinitely many .
. Then
, 
, 
is always such a triple for 
. Note that it is clear for any the last two of these are sums of squares. The hypothesis is true for 
, as 
, 
, 
. Now suppose it is true for . Then, 
, and from the inductive hypothesis both factors are sums of squares, and the product of sums of squares is also a sum of squares (since 
), which completes the proof.
works for 
. It is clear that and are expressible as a sum of two squares. Now just verify that
.
. Automatically 
. Now we need that 
is good. The quadratic residues 
are 
so any good number is either 
. So 
. So let 
. We now need that 
is good. It is well known that product of two good numbers is itself good. This follows from the identity
is good, it suffices to show that for infinitely many , 
is good. Some experimentation shows that 
works for all .
good, if 
with 
.
. Set 
, for a suitable 
. Clearly, is good.
. I claim that, 
has infinitely many solutions. (note that, this immediately clinches to be good). Indeed, this yields 
, and this is nothing but the Pell's equation, with 
. is good. Note that, 
, and thus, any prime 
must obey 
. For this, let 
is the prime decomposition of . It is well-known that, the congruence 
is solvable, for every 
prime. Thus, taking 
yields 
. Now, using Thue's theorem, there exists 
such that 
, such that 
. This yields, 
. Finally, since 
, it must follow that is a sum of two squares.
such that , , are each the sum of the squares of two integers. [Example: , , .]
, which is an example of Pell's equation. Since by definition of Pell's equation, it has infinite solutions. Therefore we're done. 
, consider 
. Then note that
Maybe it would be interesting to see this problem with the restriction that the numbers can be expressed as sum of two non-zero perfect squares. 
. This is because, take 
, then for all odd 
we get an unique .
which satisfy the above relations. Note that 
is a solution with 
.
to be a square it'll be very helpful, because if its a square then i can just simply write that term as 
and the term as 
is a triple that works 
is a triple that works which is trivially true,
for some 
and is super easy to see from the very first line
,
= 
= 
= 
)( )
)
wherein 
,
,
.
is the sum of two squares due to the Diophantus identity 
is a set of three consecutive positive integers, each the sum of two squares. This set is different from 
since 
.
And so we are done.
then 
,
,
works.