[/quote]
,
![\[y''+by'+cy=g(t)\]](http://latex.artofproblemsolving.com/7/8/8/7882dc954ab6dba2445b68d370d630ce1c66a6db.png)
with 
,
are constants has the form ![\[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]](http://latex.artofproblemsolving.com/6/3/c/63c34df1c0013bf659bc76506a93873a5f879070.png)
and use variation of parameters in each case to find the solution. Is there any short way to calculate solution?
that verfy the following equation :
be a positive integer. We say that a positive integer 
is 
is divisible by 
for all positive integers 
with 
.
is not 
is prime.
,
.
is the inverse of function 
and thus, they might intersect that 
axis (if they do, we are done). Thus: 
is the only solution.
let 
and 
be the circumcenter and the orthocenter. The point 
is the reflection of 
with respect to 
.
as 
lie on 
respectively such that 
.
be the intersection point of 
.
.
Let 
.
a grid of 9*9, we choose a little square in 
be the incenter of triangle 
.
,
,
be the orthocenters of triangles 
,
,
,
,
,
,
.
.
be 
.
,
,
,
,
,
similarly. Prove that 
,
,
are concurrent.
![$f:[1,8]\to R$](http://latex.artofproblemsolving.com/c/e/c/cec97ea0372e37977e8a77e8e77518a2344c1c43.png)
that:
that exist function 
such 
continuous on 
and discontinuous on 
Y by Mathuzb, Adventure10
Let 
be a sequence given recrusively such that 
and 
for 
Show that :
a)
is a positive integer.
b)
is a square of an integer.
Proposed by Stefan Gyurki (Matej Bel University, Banska Bystrica).
be a sequence given recrusively such that 
and 
for 
Show that :
a)
is a positive integer.b)
is a square of an integer.Proposed by Stefan Gyurki (Matej Bel University, Banska Bystrica).
This post has been edited 1 time. Last edited by XbenX, Apr 7, 2019, 1:13 PM
,
.
and hence the sequence is strictly increasing.
,
.
for 
,
and the latter concludes that 
is a positive integer for all 
,
is both an integer and a square of a rational number. Therefore, it is a square of an integer.
is a Fibonacci number.![\[ F_{4n+5} = \frac{7F_{4n+1} + \sqrt{45F_{4n+1}^2 - 36}}{2} \iff 5F_{4n+1} + 3 F_{4n} = \frac{7F_{4n+1} + \sqrt{45F_{4n+1}^2 - 36}}{2} \iff F_{4n} F_{4n+2} = F_{4n+1}^2 - 1\]](http://latex.artofproblemsolving.com/6/f/5/6f5be0db08f4c7c09410b5cfc8ddb6e071733f39.png)
which is well known or easy to verify.
which is again well known or easy to verify.
are both roots of the quadratic equation 
so that either 
or 
where the latter cannot happen because the sequence is increasing.
,
is "equivalent" to a quadratic equation 
with some additional constraint on the size of the solution (which in this case is given by the condition that we should have 
in the recurrence, not 
.
and from here it is already immediate that we have something very close to 
and examining the difference 
we immediately guess the correct recurrence.
and hence for large 
which means that we should expect a second-order linear recurrence involving the minimal polynomial of 
which of course is just 
.
.
and now it is not a big deal to come up with the idea of 
satisfying the same recurrence relation 
.
check that the condition is equivalent to 
But this is symetric with respect to 
,
are 
check that by Vieta we get that 
.
So it now follows easily by induction, as if 
is an integer, then 
is a square root of an integer that gives a rational number, so it must be an integer.
