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for which the inequality ![\[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]](http://latex.artofproblemsolving.com/4/d/b/4db42ebabc7309bba9629a9a7125be43abf113f0.png)
holds for all real numbers 
, 
, 
, and 
.
such that 
. Let excircle-
be tangent with 
at points 
respectively. Let 
and 
be points on the segment 
respectively such that 
is parallel to 
. Lastly, let 
be the circle that is externally tangent with the excircle- on point 
. Prove that 
, and concur at a point.
be an odd prime and 
is a bijection and 
is an integer such that 
Show that 
is a multiple of .
and 
denote two sequences of integers defined as follows:
Prove that, except for the "1", there is no term which occurs in both sequences.
and the same for 
then,
![\[x_n=X(-1)^n+Y2^n.\]](http://latex.artofproblemsolving.com/b/6/4/b64103ce6b36840afb72e204ce9d8bde7c3ae59e.png)
Using 
and 
yields ![\[x_n=\frac{2^n+(-1)^{n+1}}{3}.\]](http://latex.artofproblemsolving.com/0/e/9/0e91ecb368a2adce3368cc089356ac41c89c7f0a.png)
Similarly, solving for the second recurrence yields ![\[y_n=2\cdot 3^{n-1}+(-1)^n.\]](http://latex.artofproblemsolving.com/b/e/9/be9151fca1e0c704e75c410c3d6963099b9170e1.png)
So if 
then 
or 
.
or 
, then 
is the only solution, corresponding the the fact that the term 
is in both sequences. If 
, then 
. But we have 
, so ![\[2\cdot 3^n+3(-1)^n+(-1)^m\equiv 5(-1)^n+(-1)^m\pmod 8\]](http://latex.artofproblemsolving.com/c/9/6/c965e92b534b4e7746ecd90274be64ceaa3d0122.png)
which cannot be 
. Hence there are no solutions for , and the only integer in both sequences is indeed . 
cannot be equal to 
. 
.
. Now suppose this is true for some . Since 
, and 
, so adding these two gives 
, which means will always be greater than for all . 
,
and the same for 
then we would have 
.
,
and 
,
,
.
.
,
,
or 
,
.
.
.
if 
,
.
where 
and 
are roots of the equation 
.
and 
we get 
.
.
we get 
.
if 
, so we get 
.
we get 
which is not true hence , sequence 
and 
shares no common numbers other than 