be a rational number with 
and 
.
.
![\[\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}\]](http://latex.artofproblemsolving.com/d/4/5/d4511807f1e04ac2fa8aae36d65f9e0f5ba11c6f.png)
is not divisible by 
for any integer 
.
is composite, then 
.
consecutive positive integers is divisible by 
such that 
divides 
.![\[A=\frac{(m+3)^{n}+1}{3m}\]](http://latex.artofproblemsolving.com/f/5/8/f58ed8713c887510943d543c89f83640a546401f.png)
is an integer. Prove that 
is odd.
be divisible by 
.
is divisible by 
and 
be natural numbers having the property ![\[n_{2}\; \vert \; 2^{n_{1}}-1, n_{3}\; \vert \; 2^{n_{2}}-1, \cdots, n_{k}\; \vert \; 2^{n_{k-1}}-1, n_{1}\; \vert \; 2^{n_{k}}-1.\]](http://latex.artofproblemsolving.com/2/0/7/20707ef12d0e08bec8067225a9e9128bbc1d2d45.png)
Show that 
.
.
Y by Adventure10, Mango247
Let 
be a given positive integer. The sequence 
is defined as follows: 
and 
is the least positive integer which is not in 
. Show that there exist real number 
such that 
for all positive integer 
.
be a given positive integer. The sequence 
is defined as follows: 
and 
is the least positive integer which is not in 
. Show that there exist real number 
such that 
for all positive integer 
.
satisfy the equation![\[ \frac {1}{a} + \frac {1}{a + k} = 1,
\]](http://latex.artofproblemsolving.com/7/7/c/77c7cc17e891ba67c4304765e3d74b95ff4f1435.png)
![\[ a = \frac {\sqrt {k^2 + 4} - k + 2}{2}.
\]](http://latex.artofproblemsolving.com/5/0/a/50a98e04a916efbc3ed0f2c8de1ed77fdb01f2db.png)
is a positive integer, it is clear that both 
are irrational and thus produce a pair of complementary 
and 
.
is different from any of 
.
,
(a rigorous proof can be done by induction on 
