AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
that satisfy 
are positive real numbers, then
such that 
is a perfect square for all integers 
and 
.
satisfied 
Find minimum and maximum of 
such that:
with all 
is real number
+1 w
be the set of integers. Determine all functions 
such that, for all integers and , 
Proposed by Liam Baker, South Africa
be a positive integer. Find all real solutions 
to the system:![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
of positive integers is called central if for every positive integer , the arithmetic mean of the first 
terms of the sequence is equal to .
of positive integers such that for every central sequence 
there are infinitely many positive integers with 
.
are on the side 
such that 
and let 
be the projections of 
onto 
respectively. The line 
intersects 
at 
(different from point 
).
.
.
, it's easy to know that 
, from which we can get 
.
, which is the area of 
, as desired.
. We have![\begin{align*}
2[AMDN] &= 2([AMD]+[AND]) \\
&= AD(AM \sin \theta + AN \sin(A-\theta)) \\
&= AD(AF \sin \theta \cos(A-\theta) + \cos \theta \sin(A-\theta)) \\
&= AD \cdot AF \cdot \sin \theta.
\end{align*}](http://latex.artofproblemsolving.com/9/a/f/9afd7bdeab076d2d8de1ddae51b2a174a41c222e.png)
So it suffices to show that 
, but this is simply true because .
and 
Prove that
Prove that
Prove that
and 
Prove that
Now dividing both sides by 
gives the result.
Divide by 
and use 
.
Using Lemma 1: 
This is equivalent to 
Which is just Ptolemy theorem for 
.![[asy]
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dot((4.01227895316257,2.0627224343031854),linewidth(4pt) + dotstyle);
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dot((2.2642971136055023,-0.5581438324854661),linewidth(4pt) + dotstyle);
label("$E$", (2.324954239752077,-0.42176785785057574), NE * labelscalefactor);
dot((1.761920925463417,8.7237821962927),linewidth(4pt) + dotstyle);
label("$A$", (1.8299302620281561,8.851681324844241), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/a/0/b/a0b5111ee09630046735ad62d794f07764d2fd5a.png)
what is it?
.
is perpendicular to 
，
,
,
,
