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 
.
. Define the f-area of a rectangle to be f(x)*f(y) where x, y are the lengths of its sides. Since f is additive, summing the f-area over the tiling gives:
thus 
so 
and we are done.
be arbitrary real numbers (later on, 
will be those from the problem). Define 
.
of 
such that for any 
there are unique 
with 
(this set is called a special basis).
.
as in the lemma and for 
define 
.. Then 
iff 
is rational.
., then 
is rational. is rational, then 
is the unique representation of , thus 
.
we have 
and 
.
with the properties of lemma 2 and 3.![$ R=[a,b] \times [c,d]$](http://latex.artofproblemsolving.com/5/4/d/54da80bcbebfc5ad96c84b3dc4d56a0a56dc8ec4.png)
with 
and 
. It's interior is ]![$ a,b[ \times ]c,d[$](http://latex.artofproblemsolving.com/8/7/a/87ae97dbbaa73a9905ca289f2829c10aa4afa0ab.png)
and we call two rectangles distinct if their interiors are distinct as sets.
("irrational area"; but note that this can be negative).![$ R_1 = [s_1, t_1] \times [u_1, v_1] , ... , R_m = [s_m, t_m] \times [u_m, v_m]$](http://latex.artofproblemsolving.com/0/b/5/0b59dea2bfaf7f378175b8625d2f6b0cd02f4d46.png)
be pairwise distinct rectangles such that ![$ [s,t] \times [u,v] = R : = R_1 \cup ... \cup R_m$](http://latex.artofproblemsolving.com/b/e/8/be8e1747ee431063d5df36b26581680f99a59163.png)
is a rectangle again. Then 
.
. Obviously, 
is trivial. 
is easily checked with the definitions using .
of rectangles. We may assume that 
(so 
is the rectangle at the bottom left). If 
and 
we have 
and are done. W.l.o.g. we assume 
and that 
.
we are again done. Now subdivide 
into ![$ C=[s,t_1] \times [u,v]$](http://latex.artofproblemsolving.com/2/f/6/2f6547629d5b31454ca6a6071eabbf586bd2fc81.png)
and ![$ D=[t_1,t] \times [u,v]$](http://latex.artofproblemsolving.com/9/5/e/95e46d65b7ddf29d410e85b0a1d6d7992861f8b4.png)
. Then 
is union of the distinct rectangles 
, similar 
is union of 
. Now we got decompositions of 
into rectangles, and we can neglect 
and 
in these decompositions. By induction, we get that 
and 
. By using 
and 
(both distinct unions) we are done by using the case .
. Let be the big rectangle with sides 
. Then 
.
. Thus 
for all 
and by lemma 2 
is rational.
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 
.
.
and 
is arbitrarily cut into 
.
and 
for all 
.
