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.
on a plane with centers 
, respectively. 
is feet of the perpendicular from 
to 
and 
and 
are intersections of line 
with circle 
. Points 
and 
can move around the circle 
and 
, respectively. Find the maximum possible value of 
.
+3 w
, prove that 
to 
) until they have chosen seven such digits, and then consider the resulting seven-digit number by concatenating the digits in the order selected, with the seventh digit appearing last (i.e. 
). Alice wins if and only if the resulting number is the last seven decimal digits of some perfect seventh power. Please determine which player has the winning strategy.
containers placed along the track. Prove that there exists a position starting at which the car, beginning with an empty tank of gas, can complete a loop around the track without running out of gas. The tank of gas is assumed to be large enough.
such that there exists a unique integer 
such that 
with the following property:![\[ n!\mid a^n + 1
\]](http://latex.artofproblemsolving.com/e/2/3/e23d39aab685888993d355d8c93d4e6475f53487.png)
which are larger than 
and less than 
Prove that
, 
, and 
while the second equation should be 
. So I got 
, 
rather than 
, realized I had messed up somewhere (and assumed it wasn't an algebra mistake), so I gave up on the problem. 
is a polynomial such that 
and ![\[\frac{P(2x)}{P(x+1)}=8-\frac{56}{x+7}\]](http://latex.artofproblemsolving.com/1/a/b/1ab69a52821b33240a86fd8d9e64a4152262f3b8.png)
for all real 
for which both sides are defined. Find 
.
and 
Prove that
Prove that
for every real 
.
,
and 
.
,
,
,
.
for which 
.
,
.
,
,
for some polynomial 
.
,
.
is a constant polynomial. Back-substituting, we see that ![\[P(x)=(x-1)(x+1)Q(x)=(x-1)(x+1) \cdot xR(x) = c(x-1)x(x+1)\]](http://latex.artofproblemsolving.com/a/7/9/a792fa8a778493ac4a6f3dbd118f8995caec35e6.png)
for some constant 
.
and 
,
,
.
and ![\[P\left(\frac72\right) = \frac23 \cdot \frac52 \cdot \frac72 \cdot \frac92 = \frac{105}{4}.\]](http://latex.artofproblemsolving.com/f/e/6/fe6cfa5a35256c17654e66cffafd520c9dcdcaf6.png)
The answer is 
.
,
or 
.
if it is of degree 
,
.
,
,
,
and 
,
,
.
to get 
,
(hoping I didn't miscalculate)
)
.
,
gives -1 as a root. At this point plugging in 
doesn't do anything, since 
term is 0 anyways. So we have by factoring 
.
always, which means 
The RHS vanishes, hence the LHS vanishes, i.e. 
The LHS vanishes, hence the RHS vanishes, i.e. 
The LHS vanishes, hence the RHS vanishes, i.e. 
for some polynomial 
Substituting this expression into the given identity, and cancelling like terms, we obtain 
for all 
Therefore, 
It follows that 
Hence, 
,
Then
instead of 
yields 
,
are roots. By the Factor theorem, we have 
.
gives 
.
.
.
and 
to obtain 
,
.
and 
and so 
Plugging this in, 
and so 
Using the 
relation again, we get 
and plugging in 
gives the result.
.
.
gives 
and 
.
,
,
.
.
.
and 
.
and 
and so 
.
respectively.
for some polynomial 
which simplifies into 
.
.
and 
.
.
,
.
as desired.
for some polynomial 
.
.
and realize the polynomial is 
,
is the only polynomial that satisfies the given equations; obviously, it works. First, it is easy to show via substitution that 
.
for all 
.
It follows that the polynomial 
is the zero polynomial, so 
for some constant 
.
,
.
.
and 
give 
.
gives 
.
are zeroes of 
for some polynomial 
for all 
,
is a constant. Hence 
and 
,
.
,
which are obvious roots. Indeed, suppose r is a different root. If it is positive then since P(r)=0 and r-1 is not 0, P(r+1) must be 0, hence inductively there are infinite roots, contradiction. On the other hand, if r is negative by "shifting the index" to get that (r-2)P(r)=(r+1)P(r-1), since P(r)=0 and r is not -1 then P(r-1) must be 0, implying again infinite roots. Hence P(x)=a(x-1)x(x+1), and using the given condition we solve it to get that a=2/3. Now plugging in x=7/2 yields 105/4 for a final answer of 
,
hence 
.
after some simplification, giving us 
as the answer.
be a root of 
so 
or 
is also a root of 
so 
or 
is also a root of 
.
and 
so 
.
so the answer is 
where 
is the degree of 
in 
is 
and the coefficient of 
is 
,
.
.
is a root of the denominator, let 
.
is a root of the denominator, let 
.
,
.
and 
,
and 
in equation 
