AoPS Academy
consecutive positive integers is divisible by 
for which there exists an integer 
such that 
divides 
.
and be natural numbers such that ![\[A=\frac{(m+3)^{n}+1}{3m}\]](http://latex.artofproblemsolving.com/f/5/8/f58ed8713c887510943d543c89f83640a546401f.png)
is an integer. Prove that 
is odd.
and be natural numbers and let 
be divisible by 
. Show that 
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 
.
or 
then say it is the "x" part of a historic set. If it is 
or 
then say it is the "y" part of a historic set. Finally, if it is 
or 
then say it is the "z" part of a historic set. Then every nonnegative integer is part of a historic set, because if 
then 
, and 
.
and 
.
) and assign it as an x, and assign 
as a z. If 
is unassigned, then assign it as a 
, otherwise assign 
as a y. Either way, we will have made a historic set 
or 
. I will show that this algorithm is always possible to complete.
must be less than . Therefore, will be unassigned, so we can let it be 
. To show that it is possible to assign y to some value, I will show that must currently be unassigned. It cannot be an x, because it is not less than . It cannot be a y, because then its associated x value would not be less than . If it were a z, then 
must be its associated x value. However, when we assigned 
to x, we should have assigned 
as its y, a contradiction because we are assuming is unassigned.
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 
.
.
with 
is called historic if 
.