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.
a prime number with 
and 
when 
. then 
else 
.
be positive integers and 
be prime. We know that 
and 
. Prove that 
is a perfect square.
be a triangle with circumcentre 
. The points 
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
. respectively, and let 
be the circle passing through and . Suppose that the line is tangent to the circle . Prove that 
with 
such that ![\[ (a-1)(b-1)(c-1) \]](http://latex.artofproblemsolving.com/f/8/c/f8c33b59bdba1fdd12dec51f16a0b1b93e60754f.png)
is a divisor of 
is called good if there exists a permutation 
of the numbers 
, such that:
and 
have different parities for every 
is a quadratic residue modulo 
for every 
is considered a quadratic residue modulo if there exists an integer 
such that 
.
| 
with 
are positive integers and is prime (
), we get: 
be a real numbers such that 
Prove that
Where 
be non-negative real numbers. Prove that
Looking for an algebraic solution!
and 
have common external tangents 
and 
meeting at 
. Suppose touches at 
and touches at 
. A circle 
through and intersects again at 
and again at 
, such that quadrilateral 
is convex.
and 
meet at point 
, while lines 
and 
meet at point 
. Show that , , are collinear.
be an acute, scalene triangle with circumcenter 
, circumcircle 
, orthocenter 
. Line 
meets again at 
. Let 
be the midpoint of segments 
respectively. The line through and perpendicular to 
meets line at 
.
meets again at 
. Prove that the intersection of line 
and the perpendicular bisector of 
lies on the circumcircle of triangle 
. and perpendicular to 
meets line at 
. Line 
meets again at 
. Let 
be the intersection of lines 
; 
be the intersection of lines 
; 
be the intersection of lines 
. Let be the intersection of lines 
and 
be the line through and perpendicular to 
. Prove that bisects 
.
such that 
. Show that the equation 
has no integer solution.
, 
, 
, be positive numbers. Prove that
, the coefficient of the largest degree on the right side is 3^p+8^p+5^p>2*7^p, a contradiction, so our only solutions are of the form P(x)=ax^4+bx^2+c, and checking we need c=0 while all the other coefficients match up. 
, which clearly work.
denote the assertion. Clearly 
gives 
. Then from 
we see 
implying is even. Thus can be written as 
for some polynomial 
. work. It is to show that these work from just plugging.
fails. To see this we consider the leading coefficients of the polynomial. Note that we want the following to fail,
This is equivalent to showing that the leading coefficients of the following fail,
Setting 
and 
we see we need,
However it is easy to see the LHS is much greater than the RHS for 
, so it is impossible for this equation to hold. Thus we are done.
gives 
so then we can plug in 
to get that is even.
. Taking 
gives 
. Solving, we get 
.
we find that .
we find that 
, so must be even.
, 
and 
.
, and we obtain![\[ P(5k) + P(3k) + P(8k) = 2P(7k) \]](http://latex.artofproblemsolving.com/0/1/d/01dbee029b82151f3a309cf5af8f0b93de642a67.png)
Now, suppose that 
has degree 
. Then we obtain![\[ 5^n + 3^n + 8^n = 2\cdot 7^n \]](http://latex.artofproblemsolving.com/4/8/9/48968f6a13dbd3e538b437453c2935ede8599de3.png)
From Bernoulli we see that 
. Thus, 
. Taking modulo 6, we see that 
.
, as must be even.
for some reals 
and 
. To see how 
works, note that![\[ (a-b)^4 + (b-c)^4 + (c-a)^4 - 2(a+b+c)^4 = -6(ab+bc+ca)(ab+bc+ca+a^2+b^2+c^2) \]](http://latex.artofproblemsolving.com/5/5/b/55b5bb18d2789e2c80dfec95e4cb4c6687ef1716.png)
and 
works as![\[ (a-b)^2 + (b-c)^2 + (c-a)^2 - 2(a+b+c)^2 = -6(ab+bc+ca) \]](http://latex.artofproblemsolving.com/2/7/b/27bcd8e83e3c3a7fa6d78263a92599eef5beff71.png)
so indeed all polynomials of that form work. 
. Let's play with the coefficients of the polynomial, that is, if there is some 
on the left in the expression, where say is an expression of 
, then we should have the same thing on the right.
behaves: 
$$\Rightarrow \a_0=0.$$
behaves: 
is not always 
, so 
.
behaves: 
can essentially be anything, and we leave it that way.
further, but I'm so lazy, a lot of numbers appear with it, and something tells me that equality does not always happen, I thought. Then, it would be worthwhile to come up with some universal values for so that the condition 
would be satisfied at the same time, and so that the coefficients could be conveniently estimated.
looks like this: 
It is very inconvenient to disclose this whole thing with large values of 
, then let's find convenient values for 
Then, let's look at 
, I thought it would be quite convenient if 
and the value would be identical to 
for any . With the same logic, I made it so that 
, then let's see what happens to 
:
We get two roots, I see that the first one is more convenient and I take it. Let's look at our coefficient under these values: 
Let's say is not , then 
With 
and odd, the two sides will have different signs, and with 
and even, the growth will be faster on the left than on the right (it can be proved by induction that there will always be more on the left with .) Works for 
, but this does not mean that this will always be the case. If 
we have checked. Check 
, use the facts that 
, 
and make sure that the equality is identical.
and must be zero, and these two can take any value.
works where 
, 
.
denote the assertion 
.
.
. So 
.
.![\[ P\left(\frac{5c}{6}\right) + P\left(\frac{c}{2}\right) + P\left(\frac{4c}{3}\right) = 2P\left(\frac{7c}{6}\right). \]](http://latex.artofproblemsolving.com/f/f/a/ffac9e9e5f2e7743e7b1af924d8b9e8edbff4800.png)
term, we get that,![\[ a_i\left(\frac{5}{6}\right)^{2i} + a_i\left(\frac{1}{2}\right)^{2i} + a_i\left(\frac{4}{3}\right)^{2i} = 2\left(\frac{7}{6}\right)^{2i}. \]](http://latex.artofproblemsolving.com/a/3/5/a35651a4e49833d4c916728421f9a0c3d564e0a0.png)
.
, then 
. Now note that 
for all 
.
. Checking gives that 
and 
both work. So we get our desired form and we are done. 
for real numbers 
works and we want to prove its uniqueness. and 
are solutions, 
is a solution for all real numbers 
so it's enough to find all solutions of the form 
.
and 
gives us that is even
, then this obviously doesn't hold.
, then 
, which works.
, then 
, which also works.
, then 
which is a contradiction.
, then 
contradiction.
for any real number 
, which clearly works., we find that . Then, plugging 
, we have that is even.
, but this isn't possible since 
implies over reals. The next best thing would be to consolidate terms of the left hand side: 
Plugging 
gives us that 
However, we also have 
We will only use 
which when plugged into 
we get 
We will equate coefficients of 
on both sides, noting that is even. If the coefficient of 
in is 
, then we have 
Thus, or each nonnegative integer , either 
or 
This is only true for 
and , as we can test that doesn't work, and 
for 
Thus, only the and coefficients are allowed to be nonzero, thus done.
, makes it so that all terms are of the form 
This is very nice because this means that the coefficent of this only depends on the coefficient of , allowing us to easily equate coefficients. By only setting "homogeneous" constraints, like 
that I used, one can prevent the input of the polynomial from having a constant shift. In fact, 
is really the only substitution you need to solve this problem.
for real numbers works and we want to prove its uniqueness. and are solutions, is a solution for all real numbers so it's enough to find all solutions of the form . and gives us that is even, then this obviously doesn't hold., then , which works., then , which also works., then which is a contradiction., then contradiction. suffices
for real numbers 
, then it is true that the linear combination of any two solutions is still a solution, but in this case there are no solutions of the form even though there does exist solutions.
is not the only basis for real polynomials, and there are plenty of other choices, so closure under addition like this does not imply that only checking these suffices
I wish you good luck in the future!
where 
. This polynomial indeed satisfies the problem.
and 
for all 
. So we can write 
where 
. Now we put 
into the equation and after clearing the denominator we get 
for all , where 
We can check that 
for all 
. And for 
we can prove via the multinomial theorem that 
.
, which will finish the problem. Suppose ftsoc that 
and let 
. Since 
, it follows that the coefficient of 
in 
is some nonzero number times 
. Thus we must have 
, which is the desired contradiction.
be the set of solutions. We prove that 
so is an even function. is a 
-vector space
both work.
to get 
Hence, if a polynomial works, then we require 
where 
However, simply note that 
for 
. It can be shown (painfully) that 
works, thus 
works.
works, thus works, we shall show that this is the only solution.
and 
is even
, we immediately obtain![\[f(a) + f(0) + f(-a) = 2f(a) \Longrightarrow f(a) = f(-a) + f(0)\]](http://latex.artofproblemsolving.com/2/b/a/2ba188c9c1c48089b129b1a84eda8de0300f436a.png)
Since the constant terms of both sides of the equation have to be equal, it follows that , thus is even. ![\[f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c)\]](http://latex.artofproblemsolving.com/9/0/9/909c138a61d0d1f6859d7822fb7893b8db2f62ac.png)
if 
works, then 
, 
, etc, also has to work. Thus by claim 
, we simply need to show that 
fails for 
. Let us first show that 
fails fails for 
to get rid of the algebraic condition, thus our equation becomes![\[{(a-b)}^{2k} + {\left(b + \frac{ab}{a+b}\right)}^{2k} + {\left(\frac{-ab}{a+b} - a\right)}^{2k} = 2{\left(a+b-\frac{ab}{a+b}\right)}^{2k}\]](http://latex.artofproblemsolving.com/c/a/a/caa05030c62f1aba4337d3d4ac6f2a777652f548.png)
We can multiply both sides by 
to obtain![\[{(a^2-b^2)}^{2k} + {(b^2 + 2ab)}^{2k} + {(a^2 + 2ab)}^{2k} = 2{(a^2 + ab + b^2)}^{2k}\]](http://latex.artofproblemsolving.com/7/3/4/73441f6405e2b94cce970f61398dd199daf37a79.png)
Now we can substitute 
and 
which gives![\[3^{2k} + 5^{2k} + 8^{2k} = 2 \cdot 7^{2k} \Longrightarrow {\left(\frac{3}{7}\right)}^{2k} + {\left(\frac{5}{7}\right)}^{2k} + {\left(\frac{8}{7}\right)}^{2k} = 2\]](http://latex.artofproblemsolving.com/1/2/2/1228a1e5cf8f1bf0aa746fa9e853e48f0dcb71cf.png)
Since 
by the binomial theorem, we are done. works, then , , etc, also has to work. works for all 
with 
however there exists some 
such that 
, but 
fails.
denote the largest value such that 
works, but 
fails., it follows that if and 
then 
works, thus we can just get rid of these terms and only look at the function
, we also preemptively define 
denote some triple of values such that 
and for which![\[{(p-q)}^{k} + {(q-r)}^{k} + {(r-q)}^{k} \neq 2{(p+q+r)}^{k}\]](http://latex.artofproblemsolving.com/3/8/0/380f8e875791accd07888e501e00e269fe98bde3.png)
such a triple clearly exists since we assumed that doesn't fulfill the functional equation.
, since we have assumed that 
works, it follows that![\[g(p-q) + g(q-r) + g(r-p) = 2g(p+q+r)\]](http://latex.artofproblemsolving.com/6/4/0/640cd8076426eaf2a8a94c65cd989b08b4a34aaf.png)
Thus![\[a_{k} \cdot \delta = 2h(p+q+r) - h(p-q) - h(q-r) - h(r-p)\]](http://latex.artofproblemsolving.com/2/5/e/25e8535ee9a5d52bc6cbb7c5dbf1805712c18067.png)
Notice that if the functional equation holds for some , it also holds for some 
where 
is an arbitrary constant. Thus by scaling the variables by we have![\[a_{k} \cdot \delta \cdot c^{k} = 2h(cp+cq+cr) - h(cp-cq) - h(cq-cr) - h(cr-cp)\]](http://latex.artofproblemsolving.com/4/b/7/4b73951e14e1aca190bda2fe72897b2dde744943.png)
Since 
is a degree 
polynomial at best, it follows that the right hand side grows by at most 
, whereas the left hand side grows by 
., the left hand side cannot possibly equal the right hand side. Thus we are done.
. Then, if 
,
works since 
, and there are no other solutions. is even and of degree 4, let 
. Plugging in, 
.
.
be real numbers such that 
.
.
with real coefficients such that for all reals 
we have the following relations![\[ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c). \]](http://latex.artofproblemsolving.com/4/4/9/4491a53f9c6f96c18e42eb984b8e83566140bfdb.png)
is in the image of f (I mean, for any such a,b,c there exists a number 
such that that number equals f(g(a,b,c))).
,
whenever 
divides 
.
and 
if 
are linearly independent, so each coefficient 
must be zero.
or 
,
,
.
.
.
,
for all even 
.
for all 
(so that 
)
.
satisfying the functional equation
for all 
.
.
,
or letting 
,
.
appears with multiplicity at least 
on the 
,
as well.
,
,
so that 
(since 
are the only possibly solutions. A quick check shows that these solutions are indeed valid.
.
,
.