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for which 
and 
are both perfect squares.
and 
and so 
Plugging this in, 
and so 
Using the 
relation again, we get 
and plugging in 
gives the result.
gives 
.
gives 
.
gives 
and 
.
gives 
.
, then 
, a contradiction as it implies 
is the zero polynomial. is a cubic polynomial with roots 
.
. 
.
and 
. We have 
and 
and so 
.
to get to get 
to get 
to get 
and .
for some polynomial 
. We plug in values of 
and 
to confirm that 
. Therefore the answer is 
combined with something in terms of 
having infinitely many roots.
is the only polynomial that satisfies the given equations; obviously, it works. First, it is easy to show via substitution that 
. Next, let 
for all 
. Plugging into the given equation, 
It follows that the polynomial 
is the zero polynomial, so 
is constant.
for some constant 
. Utilizing the second constraint, 
, so the answer is 
. Extract 
.
and 
give 
. Then plugging in 
gives 
. Since 
are zeroes of , we can write 
for some polynomial 
. Plugging this into the functional equation gives 
for all 
, so 
is a constant. Hence 
and 
, so 
. We then have that 
, so the requested sum is 
.
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 
be a root of .
so 
or 
is also a root of .
so 
or 
is also a root of .
.
and 
so 
.
so the answer is 
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 
.
where 
is the degree of . Then, the coefficient of 
in 
is 
and the coefficient of in 
is 
, so 
.
gives , so let 
. Then, since 
is a root of the denominator, let 
. Then, since 
is a root of the denominator, let 
.
, so 
.
for every real 
.
,
and 
are relatively prime positive integers. Find 
.
,
.
,
.
,
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 
,
)
,
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
,
,
while the second equation should be 
.
,
rather than 
,
instead of 
yields 
,
are roots. By the Factor theorem, we have 
.
gives 
.
.
.
,
.
for some polynomial 
which simplifies into 
.
.
.
.
,
.
and realize the polynomial is 
,
,
hence 
.
after some simplification, giving us 
as the answer.
and 
,
and 
in equation 
