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.
with 
. Show that: 
.
be positive integers such that 
and 
. Let 
be the set of values attained by 
as 
runs through the positive integers. Show that is the set of all positive divisors of some positive integer.
be a set of integers (not necessarily positive) such that
with 
;
and 
are elements of (possibly equal), then 
also belongs to . is the set of all integers.
the set of all positive integers. Find all functions 
such that for all positive integers 
and , the integer 
is nonzero and divides 
.
of positive integers such that 
is divisible by 
.
be an odd integer. Determine all functions 
from the set of integers to itself, such that for all integers and the difference 
divides 
for which there exist infinitely many positive integers 
such that ![\[ \frac{a^m+a-1}{a^n+a^2-1} \]](http://latex.artofproblemsolving.com/2/8/a/28a68e409451bbdcaee7c1286341f6aeea9ffdad.png)
is itself an integer.
such that ![\[ a^2b-1 \mid ab^3-1. \]](http://latex.artofproblemsolving.com/7/3/5/735fd61de9a18113846ce2048a00889dcc5ce8a4.png)
, 
be positive integers such that 
is a multiple of 
for all positive integers . Prove that 
. for which ![$\dfrac{n^2+1}{[\sqrt{n}]^2+2}$](http://latex.artofproblemsolving.com/7/e/d/7ed6c67114c0c7d262f67e5b7a733daab7997415.png)
is an integer. Here ![$[r]$](http://latex.artofproblemsolving.com/2/f/c/2fcabe2c2b8ebe2c428830acb9874cb8a8df790f.png)
denotes the greatest integer less than or equal to 
.
of positive integers such that 
is multiple of 
and 
is multiple of 
.
, we define a sequence by 
, 
, and 
for 
. Prove that if 
for some , then the sequence is periodic.
contains only one set of three distinct points, 
and 
which are the vertices of an equilateral triangle.
be the set of all triples 
of positive integers for which there exist triangles with side lengths 
Express ![\[\sum_{(a,b,c)\in T}\frac{2^a}{3^b5^c}\]](http://latex.artofproblemsolving.com/9/f/6/9f6e0cee87e5b6958e143ff23b887bcd876f6cba.png)
as a rational number in lowest terms.
. The general setting with arbitrary starting values leads to "essentially" the same problem, but restricting those values to 1 helps to avoid unnecessary variables.
, so that 
will be the biggest prime divisor of 
. In order to gain some familiarity with the situation, the following table provides the relevant values for 
: 
2 2 2/3 4 2 1/2 6 3 3/5 9 3 1/2 12 3 3/7 15 5 5/8 20 5 5/9 25 5 1/2 30 5 5/11 35 7 7/12 42 7 7/13 49 7 1/2 56 7 7/15 63 7 7/16 70 7 7/17 77 11 11/18 88 11 11/19 99 11 11/20 110 11 11/21 121 11 1/2 132 11 11/23 143 13 13/24 156 13 13/25 169 13 1/2 182 13 13/27 195 13 13/28 208 13 13/29 221 17 17/30
-column, to the third column as the -column, and to the respective entries as -values and -values.
is 
, and we will now examine that hypothesis.. In lines 12 to 17, for example, we consistently have the value 7 in the -column. We will now investigate these sections a little closer, focussing on the values of and . For that purpose, we define 
to be the 
-th prime number (in their natural increasing order), i.e. 
, 
etc.. We observe that the -value and the -value, that is 6 and 3, can be written as 
and for 
. Plainly speaking, the -value is the product of the corresponding prime in the -column and the previous prime. We will show that this is no coincidence for the first line of such a section.-value 3, we do indeed have and in those two columns. By definition of , the value in the -column in the next line is 
which is the same as 
. If the value in the -column of that line does not change, is added to the -value in the following line once again, resulting in the value 
there, and as long as nothing changes in the -column, the values in the -column will be of the form 
, 
in the following lines.-column will change once we reach the smallest 
such that has a prime factor 
larger than . As two different prime numbers are always relatively prime, this is equivalent to the fact that divides 
. We start with 
, when is smaller than and also smaller than any candidate prime number (which must even be greater than ). Clearly, the first such that has a prime divisor greater than is the one with 
, so that our new prime number will be 
, which then does not only divide , but will be equal to it.
lines (as that was the crucial value of which initiated a change in the -column). The last line of that section will have the value 
in the -column and in the -column, and the new section will therefore begin with a line that has 
in the -column and 
in the -column. In particular, this shows that the values in the -column run through all the prime numbers in a monotonously increasing way.-column in terms of the sequence 
. Now we have to consider them as actual elements of the sequence 
, which, loosely speaking, means that we have to find a relation between those values and the line number.-column, the prime number prevails for exactly lines. Thus the prime number 
, which appears for the first time in line 5, is succeeded by 
in line 
. By induction, the prime number (for some integer ) will appear in the -column for the first time in line
and this telescoping sum is the same as
-value does not change within a section (by definition of a section), we can conclude that, at the end of the section, which comes 
lines later, we have 
are clearly strictly decreasing, as it is only the index which changes. Therefore, to establish the limit of , which is the ultimate objective of this text, it suffices to focus on the values of at the beginning and at the end of each section. If the values of the subsequence at the beginning, i.e.
and at the end, i.e.
converge to the same limit, the entire sequence 
will converge to that same limit by the squeezing theorem. There is the (still open) conjecture that there are infinitely many twin primes. If we assume that the conjecture is true, this would easily show that the only possible limit of our two subsequences from above, and therefore the whole sequence, is indeed . is needed. I am somewhat hesitant to proceed here, however, as I feel that this is beyond what is reasonable for a problem from a Math olympiad. To me, the attraction of this problem lies in uncovering the more elementary results from above. If others can produce an elementary proof of the limit statement, though, I am very interested in hearing about it.
is the greatest prime divisor of 
.
and 
?