such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
and 
.
and 
Prove that
and 
Prove that
Prove that 
be positive integers such that 
.
.
and 
.
.
Y by siavosh, Adventure10, Mango247
A quick solution:
Let R be the foot of the perpend. from X to BC. Let's assume Q and R are in the interior of the segms AC and BC (respectively) and P in the ext of AD. P, R, Q are colinear (Simson's thm). PQ tangent to circle XRD iff XRQ=XDR iff Pi-XCA=XDR iff XBA=XDR=XDC=ADB iff XBC+ABC=ADB=DAC+ACB iff XAC+ABC=DAC+ACD iff ABC=ACD=ACB iff AB=AC. It's the same for all the other cases.
Let R be the foot of the perpend. from X to BC. Let's assume Q and R are in the interior of the segms AC and BC (respectively) and P in the ext of AD. P, R, Q are colinear (Simson's thm). PQ tangent to circle XRD iff XRQ=XDR iff Pi-XCA=XDR iff XBA=XDR=XDC=ADB iff XBC+ABC=ADB=DAC+ACB iff XAC+ABC=DAC+ACD iff ABC=ACD=ACB iff AB=AC. It's the same for all the other cases.
be a triangle. 
is a point on the side 
.
meets the circumcircle again at 
.
is the foot of the perpendicular from 
,
is the foot of the perpendicular from 
.
is a tangent to the circle on diameter 
if and only if 
.
![[asy]
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be the foot of perpendicular from 
on 
First, note that 
are collinear because they all lie on the Simson line of 
is tangent to 
iff 
However, note that 
Also, since 
and 
quadrilateral 
is cyclic, so 
Again, from cyclic quadrilateral 
we deduce that![\[\angle QBX = 180^{\circ} - \angle ABX = \angle ACX = \angle BCA + \angle BCX = \angle BCA + \angle BAD.\]](http://latex.artofproblemsolving.com/0/f/4/0f456b24bceec454208ab796e4a72d6be9b56b86.png)
Hence, we need 
as desired. 
,
.
with 
as 
.
is tangent to 
,
.
,
lie on the Simson Line, 
at 
must hold. Note that 
,
.
,
,
is cyclic, 
.
,
.
is cyclic, 
,
,
.
,
.
,
.
,
,
