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 
.
with 
, 
subject to some conditions like 
(cubic residue symbol) for all 
, 
which occur naturally in the proof using cubic reciprocity: sketch: by cubic reciprocity, ![$ [\frac{y + b\sqrt{-3}}{t}] = 1$](http://latex.artofproblemsolving.com/5/1/9/519920eb98ece0f0070e4f647cff6e0f3b75be41.png)
, which is absurd by factoring 
and using the cubic supplementary law.
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 
.
.
has no solution in integers if 
,
,
does not divide 
,
is odd if 
has a solution in integers 
if 
and 
.