straightforward solution + 2 pages of calculations = bad IMO problem. Darij

In my opinion the first day characteristic was uglyness. And this problem is the best example. Of course, problem 3 is quite nice and difficult, but I bet another combinatorics problem could have been given. Again, just my opinion.

darij grinberg wrote: straightforward solution + 2 pages of calculations = bad IMO problem. And I say this inspite of the fact that I have solved it correctly and many others have made errors dg

Why do you say 2 pages of computations? I might be wrong, but doesn't it reduce immediately to an exponential equation which has bounded solution ( i mean, you only need to see if 6 is a solution, which yes requires computation) by 7. But probably I'm wrong. Anyway, my opinion is that this problem is very idiot. Sorry...

I think a much interesting problem is the following: find all polynomials f such that for any such a,b,c like in the official problem, the number $\frac{f(a-b)+f(b-c)+f(c-a)}{2} $ is in the image of f (I mean, for any such a,b,c there exists a number $ g(a,b,c) $ such that that number equals f(g(a,b,c))).

Actually the answer is $P\left( x\right) = kx^4 + lx^4$, with constant real k and l. But in order to prove that $x^4$ works, one needs some rather messy calculations Darij

you're actually right harazi(ok, i needn't tell you about mathlinks, anyway). but some idiots like me are too dumb to check that it works for n=4. hope they don't cancel too many points.

A reasonably clean solution: We want to find all polynomials f such that $f(a-b)+f(b-c)+f(c-a)=2f(a+b+c)$ whenever $ab+bc+ca=0$. Equivalently, $ab+bc+ca$ divides $f(a-b)+f(b-c)+f(c-a)-2f(a+b+c)$. Let $f(x)=r_nx^n+r_{n-1}x^{n-1}+...+r_0$ and $a'=ta,b'=tb,c'=tc$ Then $a'b'+b'c'+c'a'=0$ if $ab+bc+ca=0$ so $t^nr_n((a-b)^n+(b-c)^n+(c-a)^n-2(a+b+c)^n)+...+r_0(1+1+1-2)=0$ The functions of t $t^n,t^{n-1},...,1$ are linearly independent, so each coefficient $r_k((a-b)^k+(b-c)^k+(c-a)^k-2(a+b+c)^k)$ must be zero. This means that either $r_k=0$ or $ab+bc+ca$ divides $(a-b)^k+(b-c)^k+(c-a)^k-2(a+b+c)^k$ For $k=2$, $(a-b)^2+(b-c)^2+(c-a)^2-2(a+b+c)^2=-4(ab+bc+ca)$ For $k=4$, $(a-b)^4+(b-c)^4+(c-a)^4-2(a+b+c)^4= -6(ab+bc+ca)(2a^2+2b^2+2c^2+ab+bc+ca)$ Consider the case $a=2,b=2,c=-1$. Then $(-3)^k+(-3)^k=2*3^k$ This forces k to be even. Now take $a=6,b=3,c=-2$. This gives $3^k+5^k+(-8)^k=2*7^k$. Since $8^6>2*7^6$, $3^k+5^k+(-8)^k>2*7^k$ for all even $k>4$. By previous arguments, we must have $f(x)=dx^4+ex^2$

This problem becomes very simple once we get rid of all the dependencies (after milking them for all they're worth): $a=b=c=0: f(0)=0$ $b=c=0: f(a) = f(-a)$ for all $a$. Therefore, all of $f$'s exponents have even degree. Then making the substitutions $x=a-b, y=b-c, z=c-a, w=a+b+c$ (so that $x+y+z=0$), the given condition becomes $2w^2 = x^2+y^2+z^2$. Therefore, the original problem is completely equivalent to finding all even polynomials $f$ with $f(0) = 0$ satisfying the functional equation $f(x)+f(y)+f(x+y) = 2f(\sqrt{x^2+xy+y^2})$ for all $x,y \in R$. Suppose $deg f = 2n$. Equating terms of degree $2n$, we find that $x^{2n} + y^{2n} + (x+y)^{2n} = 2(x^2+xy+y^2)^n$ or letting $y=tx$, $1+t^{2n}+(1+t)^{2n} = 2(1+t+t^2)^n$. Assume that $n \geq 3$. Then the root $\omega = \frac{-1+\sqrt{3} i}{2}$ appears with multiplicity at least $3$ on the $RHS$, and hence on the $LHS$ as well. However, letting $g(t)=LHS$, $0=g''(\omega) = 2n(2n-1)(\omega^{2n-2}+(1+w)^{2n-2}) = 2n(2n-1)(\omega^{2n-2}+(-\omega^2)^{2n-2}) = 2n(2n-1)\omega^{2n-2}(1+w^{2n-2}) \not= 0$, a contradiction. It follows that $deg f \leq 4$ so that $f(x) = kx^4 + lx^2$ (since $f(0)=0$) for $k,l \in R$ are the only possibly solutions. A quick check shows that these solutions are indeed valid. The fact that I found this problem simple supports your opinions, namely that this problem is relatively simple. However, I disagree that this problem is ugly. I think the concepts behind it are reasonably nice, and appropriate. But then again, this is all a matter of perspective. If you, Darij, spent 2 pages tediously computing this answer, I understand why you're bitter. But it's really not that tedious, if you check things the right way. When you reduce this to 2 variables instead of 3, it's very simple to check manually (or doing a root check the way I did above...the above method (with slightly more explanation) shows necessity AND sufficiency). Finally, I'm not sure how they mark these, but wouldn't you be safe just saying that "plugging this back into the functional equation, we find that these solutions indeed satisfy it"? That way, you avoid the hassle of computing the thing over 2 pages, saving lots of time and effort. ~Mike

If a=b=c you get f(0)=0. With a=b=0 we get f(c)=f(-c), then f(x)=P(x^2) for some P(x).(P(0)=0). Then put x=a-b=b-c (you can easy check that the system have real solution for infinite values of x) then the identity becames 2P(x^2)+P(4x^2)=2P(3x^2)..... looking at the leading coefficient we find that 2+4^n=2*3^n. But 4^n>2*3^n for n>2 hence f(x)= ax^2+bx^4 as we want.

A much less elegant solution : (1) we have f(0)=0, f is even by trivial substituions (2) Define g(x1,x2,k), where x1,x2 are reals and k is an even integer: g(x1,x2,k) = (x1-x2)^k + (x2+x3)^k + (x1+x3)^k - (x1+x2-x3)^k, where x3 = - (x1*x2)/(x1 + x2). since f is even, let the coefficients of x^2,x^4 ... x^(2n) be a(2),a(4) ... a(2n). it follows from the given condition that, for all reals x1,x2: sum, from k=1 to n of a(2k)*g(x1,x2,2k) = 0 ... (*) now g(x1,-3*x1,2k) = x1^2k ( 4^(2k) + (3/2)^(2k) + (5/2)^(2k) - 2*(7/2)^2k) (where x1 is not 0). now after some computations we have b(2k) = 4^2k + (3/2)^2k + (5/2)^2k - 2*(7/2)^2k is strictly positive for k>2, and zero for k=1,2. (for k=1,2 we just verify. for k>=3, 8^k > 2*7^k) using that in (*), and setting for convenience c(2k)=b(2k)*a(2k), sum of, from k=3 to n, x^(2k) * c(2k) = 0 for all x. this implies c(2k) = 0 for all k>3, and since b(2k) is non-zero for thos values, a(2k)=0 for all k>3 -- > f is of the form a*x^2 + b*x^4. now we verify that it works ... i totally agree that this problem isn't a very good IMO problem, despite some of the nicer solutions that it does have. It doesn't really use any clever ideas at all, and is a bit too simple for q2/5 level problems.

Indeed my solution was quite similar to all of your ones. Michael Lipnowski wrote: Finally, I'm not sure how they mark these, but wouldn't you be safe just saying that "plugging this back into the functional equation, we find that these solutions indeed satisfy it"? That way, you avoid the hassle of computing the thing over 2 pages, saving lots of time and effort. Well, maybe they wouldn't like this... the calculation is really heavy if you do not have the right idea, so I think one would lose some points with this. darij

Anyway, a really cool solution is the following: just take instead of a,b,c the numbers 6x, 3x, -2x. Then identify the coefficients and you will find a simple exponential equation. That's all!

It is really nice and natural idea. In sln of jmery it was mentioned in other form. I think it is shortest way to find that $degf \le 4$. The rest is so trivial. Namdung

Anyway, I find much more interesting and difficult to find the derivable functions that verify those conditions. I solved it and it really is vry very nice. Too bad such a question couldn't have been given in an IMO.

This is my thought: (1) If f(x) satisfies the condition, then kf(x) also satisfies the condition. (2) If f(x) and g(x) satisfy the condition, then f(x) + g(x) also satisfies the condition. (3) Let b = c = 0, then f(a) = f(-a), f(x) is even function. (4) Let a = 6, b = 3, c = -2, find n <= 4. (5) If f(x) is constant, then f(x) = 0. (6) f(x) = x^4 and f(x) = x^2 satisfy the condition. Then f(x) = kx^4 + mx^2.

My solution is similar to Harazi. Let a=b=c=0. We observe that P(0)=0. Let a=b=x, c=-x/2. We observe that all powers of P(x) are even. Let a=-2x, b=3x, c=6x. We have P(-3x)+P(-5x)+P(8x)=2P(7x). In other words, the polynomial Q(x)=P(-3x)+P(-5x)+P(8x)-2P(7x)=0 for all real number x. Therefore all coefficients of Q(x) must be 0, from which we deduce that $P(x)=sx^4+tx^2$. PS: Here's a confusing solution, I'm sure the official solution do not have this answer: let P(x)=$\infty$ for all x, then obviously P(a-b)+P(b-c)+P(c-a)=2P(a+b+c)=$\infty$ for all real numbers a,b,c. Try to prove this solution is false. I think it's better if the problem statement is as follow: "Find all polynomials with FINITE POWERS AND FINITE COEFFICIENTS such that ..."

mozilla wrote: P(x)=$\infty[$quote] is this a polynomial ??

also the creator of this problem is hojoo lee

It's $P(x)$ instead of $f(x)$ in OTIS, but same thing

We claim the answer is $f(x) = mx^4 + nx^2$ for any reals $m, n$. First, we will show that these work. Note that if $p(x)$ and $q(x)$ are two polynomials satisfying the given equation, then $\alpha p(x) + \beta q(x)$ is also a solution for any reals $\alpha, \beta$. Therefore, it suffices to show that $x^2$ and $x^4$ work. Starting with $x^2$, we see that $$\begin{aligned} (a - b)^2 + (b - c)^2 + (c - a)^2 &= 2(a^2 + b^2 + c^2) - 2(ab + bc + ca) \\ &= 2(a^2 + b^2 + c^2 + 2(ab + bc + ca)) \\ &= 2(a + b + c)^2. \end{aligned}$$To see why $x^4$ works, note that $$\begin{aligned} (a - b)^4 + (b - c)^4 + (c - a)^4 &= 2(a^4 + b^4 + c^4) - 4\sum_{\text{sym}}a^3b + 6\sum_{\text{cyc}} a^2b^2 \\ &= 2(a^4 + b^4 + c^4) - 4\sum_{\text{sym}}a^3b + 6\sum_{\text{cyc}}a^2b^2 - 2(ab + bc + ca)(ab + bc + ca - 2(a^2 + b^2 + c^2)) \\ &= 2(a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2)) \\ &= 2(a^2 + b^2 + c^2)^2 \\ &= 2(a^2 + b^2 + c^2 + 2(ab + bc + ca))^2 \\ &= 2(a + b + c)^4. \end{aligned} $$To see that no other solutions work, let $P(a, b, c)$ denote the original assertion. Then $P(0, 0, 0)$ gives $f(0) = 0$, and $P(x, 0, 0)$ gives $f(x) + f(-x) = 2f(x) \iff f(x) = -f(x)$, so $f$ only has terms of even degree. Now, let $n$ be the degree of $f$, and $a_n$ be the leading coefficient. Then from $P(-2, 3, 6)$, since $n$ is even we get $$ \begin{aligned} a_n(3^n + 5^n + 8^n) &= 2a_n(7^n) \\ \iff \left(\frac{3}{7}\right)^n + \left(\frac{5}{7}\right)^n + \left(\frac{8}{7}\right)^n &= 2 \end{aligned}$$But for $n \ge 6$ we have $(\tfrac{8}{7})^n \ge 2$, so we must have $n = 2$ or $n = 4$. Since $f(0) = 0$, $f$ has no constant term, so the only solutions must be of the form $f(x) = mx^4 + nx^2$ for real $m, n$, as desired.

In OTIS where f(x) is P(x). Notice 0,0,0 yields P(0)=0, whence a,0,0 gives P(a)+P(-a)=2P(a), so P is even. 3x,6x,-2x gives P(-3x)+P(8x)+P(-5x)=2P(7x); It's obvious that for P having degree $2p\ge 6$, 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. $\blacksquare$

Kinda disappointed this boiled down to guess and check, but okay. We claim the only solutions are of the form $f(x) = mx^4 + nx^2$, which clearly work. Let $Q(a, b, c)$ denote the assertion. Clearly $Q(0, 0, 0)$ gives $P(0) = 0$. Then from $Q(x, 0, 0)$ we see $P(x) = P(-x)$ implying $P$ is even. Thus $P$ can be written as $G(x^2) = P(x)$ for some polynomial $G(x)$.

to show that these work from just plugging. Now we claim if $\text{deg}(P) > 6$ fails. To see this we consider the leading coefficients of the polynomial. Note that we want the following to fail, \begin{align*} (a-b)^{2n} + \left(b + \frac{ab}{a+b}\right)^{2n} + \left(a + \frac{ab}{a+b}\right)^{2n} = 2\left(a + b - \frac{ab}{a+b}\right)^{2n} \end{align*}This is equivalent to showing that the leading coefficients of the following fail, \begin{align*} (a^2 - b^2)^{2n} + \left(2ab + b^2 \right)^{2n} + \left(2ab + a^2 \right)^{2n} = 2\left(a^2 + b^2 + ab\right)^{2n} \end{align*}Setting $a = 2x$ and $b = x$ we see we need, \begin{align*} (3x)^{2n} + (5x)^{2n} + (8x)^{2n} = 2 \cdot (7x)^{2n} \end{align*}However it is easy to see the LHS is much greater than the RHS for $n \geq 3$, so it is impossible for this equation to hold. Thus we are done.

Notice $(a,b,c)=(0,0,0)$ gives $P(0)=0$ so then we can plug in $(a,b,c)=(x,0,0)$ to get that $P$ is even. Set $P(x) = a_{2n}x^{2n} + a_{2n-2}x^{2n-2} + \cdots + a_2x^2 + a_0$. Taking $(a,b,c) = (3x,6x,-2x)$ gives $\deg P <6$. Solving, we get $\boxed{P(x)=cx^4+dx^2, \ c,d \in \mathbb{R}}$.

Taking $a=b=c=0$ we find that $P(0) = 0$. Taking $a=b=0$ we find that $P(-c) = P(c)$, so $P$ must be even. Now, let $a = -2k$, $b = 3k$ and $c = 6k$. Clearly this satisfies $ab+bc+ca=0$, and we obtain \[ P(5k) + P(3k) + P(8k) = 2P(7k) \]Now, suppose that $P(x)$ has degree $n$. Then we obtain \[ 5^n + 3^n + 8^n = 2\cdot 7^n \]From Bernoulli we see that $\frac{8^7}{7^7} \geq 2$. Thus, $n \leq 7$. Taking modulo 6, we see that $n \neq 6$. Therefore, the maximal degree is $4$, as $n$ must be even. Then, we find that $P(x) = k_1x^4 + k_2x^2$ for some reals $k_1$ and $k_2$. To see how $x^4$ works, note that \[ (a-b)^4 + (b-c)^4 + (c-a)^4 - 2(a+b+c)^4 = -6(ab+bc+ca)(ab+bc+ca+a^2+b^2+c^2) \]and $x^2$ works as \[ (a-b)^2 + (b-c)^2 + (c-a)^2 - 2(a+b+c)^2 = -6(ab+bc+ca) \]so indeed all polynomials of that form work. $\blacksquare$

Let our polynomial be of the form $P(x)=a_n \cdot x^n+a_{n-1} \cdot x^{n-1}+...+a_1\cdot x+a_0$. Let's play with the coefficients of the polynomial, that is, if there is some $c_i*A$ on the left in the expression, where say $A$ is an expression of $a,b,c$, then we should have the same thing on the right. Let's look at how the coefficient $a_0$ behaves: $$a_0+a_0+a_0=2a_0,$$$$\Rightarrow \a_0=0.$$ Let's see how $a_1$ behaves: $$a_1((a-b)+(b-c)+(c-a))=2a_1(a+b+c),$$$$\Rightarrow 0=a_1(a+b+c).$$ $a+b+c$ is not always $0$, so $a_1=0$. And how $a_2$ behaves: $$a_2((a-b)^2+(b-c)^2+(c-a)^2)=2a_2(a+b+c)^2,$$$$\Rightarrow a_2(2a^2+2b^2+2c^2)=a_2(2a^2+2b^2+2c^2).$$ Everything works, so we can say that $a_2$ can essentially be anything, and we leave it that way. Of course, I would like to work with $a_3$ 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 $a,b,c$ so that the condition $ab+bc+ac=0$ would be satisfied at the same time, and so that the coefficients could be conveniently estimated. The equality of the coefficient $a_i$ looks like this: $$a_i((a-b)^i+(b-c)^i+(c-a)^i)=2a_i(a+b+c)^i.$$It is very inconvenient to disclose this whole thing with large values of $i$, then let's find convenient values for $a,b,c.$ Then, let's look at $(a-b)$, I thought it would be quite convenient if $a=b+1$ and the value would be identical to $1$ for any $i$. With the same logic, I made it so that $c=b+2$, then let's see what happens to $ab+bc+ac$: $$b(b+1)+b(b+2)+(b+1)(b+2)=0,$$$$3b^2+6b+2=0,$$$$ \Rightarrow b_1=\frac{1}{\sqrt{3}}-1, \ b_2=-\frac{1}{\sqrt{3}}-1.$$We get two roots, I see that the first one is more convenient and I take it. Let's look at our coefficient $a_i$ under these values: $$a_i((\frac{1}{\sqrt{3}}-(\frac{1}{\sqrt{3}}-1))^i+((\frac{1}{\sqrt{3}}-1)-(\frac{1}{\sqrt{3}}+1))^i+((\frac{1}{\sqrt{3}}+1)-\frac{1}{\sqrt{3}})^i)=2a_i(\sqrt{3})^i,$$$$\Rightarrow a_i((-2)^i+2)=2a_i(\sqrt{3})^i.$$Let's say $a_i$ is not $0$, then $(-2)^i+2=2(\sqrt{3})^i.$ With $i\geq 3$ and odd, the two sides will have different signs, and with $i\geq 6$ and $i$ 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 $i\geq 6$.) Works for $i=2.4$, but this does not mean that this will always be the case. If $i=2$ we have checked. Check $i=4$, use the facts that $(a+b+c)^2=a^2+b^2+c^2$, $(ab+bc+ac)^2=0,$ $(a^2+b^2+c^2)(ab+bc+ac)=0$ and make sure that the equality is identical. Accordingly, any coefficient except $a_4$ and $a_2$ must be zero, and these two can take any value. Answer: $\boxed{P(x)=a_4\cdot x^4+a_2 \cdot x^2.}$

$P \equiv kx^4 + lx^2$ works where $k$, $l\in \mathbb R$. Let $Q(a,b,c)$ denote the assertion $P(a-b)+P(b-c)+P(c-a) = 2P(a+b+c)$. $Q(0,0,0)\implies P(0) = 0$. \begin{align*} Q\left(c,c,-\frac{c}{2}\right) &\implies P(0) + P\left(\frac{3c}{2}\right) + P\left(\frac{-3c}{2}\right) = 2 P\left(\frac{3c}{2}\right)\\ &\implies P\left(\frac{-3c}{2}\right)=P\left(\frac{3c}{2}\right)\\ &\implies P(-x) = P(x)\\ &\implies P(x) \text{ is even} .\end{align*} All the terms are of even degree. $P(x) = a_nx^{2n} + \cdots + a_1x^2 + a_0x^0$. So $P(0) = 0 \implies a_0 = 0$. Now $Q\left(\frac{-c}{3},\frac{c}{2},c\right) \implies P\left(\frac{-5c}{6}\right) + P\left(\frac{-c}{2}\right) + P\left(\frac{4c}{3}\right) = 2P\left(\frac{7c}{6}\right)$. So, we have, \[ P\left(\frac{5c}{6}\right) + P\left(\frac{c}{2}\right) + P\left(\frac{4c}{3}\right) = 2P\left(\frac{7c}{6}\right). \] Now comparing the coefficients of the $c^{2i}$ term, we get that, \[ a_i\left(\frac{5}{6}\right)^{2i} + a_i\left(\frac{1}{2}\right)^{2i} + a_i\left(\frac{4}{3}\right)^{2i} = 2\left(\frac{7}{6}\right)^{2i}. \]$\implies a_i(5^{2i} + 3^{2i} + 8^{2i}) = a_i(2\cdot 7^2i)$. If $a_i\neq 0$, then $25^i + 9^i + 64^i = 2\cdot 49^i$. Now note that $64^i > 2\cdot 49^i$ for all $i\ge 3$. Thus we must have $i\le 2$. Checking gives that $i=1$ and $2$ both work. So we get our desired form and we are done.

We claim our solutions are of the form $\boxed{P(x)=px^4+qx^2, \quad p,q \in \mathbb{R}}$, which can be tested to work. Denote the assertion as $A(a,b,c)$. Then \begin{align*} A(0,0,0) &\implies P(0)=0 \\ A(a,0,0) &\implies P(a)=P(-a) \implies P \text{ even.} \\ A(1, 1-\sqrt 3, 1+\sqrt 3) &\implies (\sqrt 3)^{2k} + (-2\sqrt 3)^{2k} + (\sqrt 3)^{2k} = 2 \cdot 3^{2k} \\ &\implies 4^k+2 = 2 \cdot 3^k \\ &\implies k=1,2. \end{align*} Hence the only possible degrees of nonzero terms of $P$ are 2 and 4, as desired. $\blacksquare$

We can check that $P(x)=mx^4+nx^2$ for real numbers $m,n$ works and we want to prove its uniqueness. Note that if $P$ and $Q$ are solutions, $aP+bQ$ is a solution for all real numbers $a,b$ so it's enough to find all solutions of the form $x^n$.$P(0,0,0)$ and $P(0,0,x)$ gives us that $n$ is even $$P(6,3,-2) \Rightarrow 3^n+5^n+(-8)^n=2 \cdot 7^n$$ If $n=0$, then this obviously doesn't hold. If $n=2$, then $9+25+64= 98 =2 \cdot 49$, which works. If $n=4$, then $81+625+4096 = 4802 = 2 \cdot 2401 $, which also works. If $n=6$, then $$3^6+5^6+8^6>8^6=(7+1)^6=7^6+\frac{6}{7} \cdot 7^6 + \frac{15}{49} \cdot 7^6 = (2+\frac{8}{49}) \cdot 7^6>2 \cdot 7^6,$$which is a contradiction. If $n \ge 8$, then $$3^n+5^n+8^n>8^n=(7+1)^n>7^n+n \cdot 7^{n-1} > 2 \cdot 7^n,$$contradiction. $$\mathbb{Q.E.D.}$$

We claim that the answer is $P(x)=rx^4+sx^2$ for any real number $r,s$, which clearly works. By plugging $a=b=c=0$, we find that $P(0)=0$. Then, plugging $b=c=0$, we have that $P$ is even. We would like to cancel the right side by setting $a+b+c=0$, but this isn't possible since $a+b+c=0,ab+ac+bc=0$ implies $a=b=c=0$ over reals. The next best thing would be to consolidate terms of the left hand side: $$a-b=b-c$$$$c=2b-a.$$Plugging $c=2b-a$ gives us that $$2P(a-b)+P(2b-2a)=2P(3b).$$However, we also have $$ab+ac+bc=0$$$$ab+(2b-a)(a+b)=0$$$$a^2-2ab-2b^2=0$$$$a=b(1\pm \sqrt{3}).$$We will only use $a=b(1+\sqrt{3}),$ which when plugged into $$2P(a-b)+P(2b-2a)=2P(3b),$$we get $$2P(b\sqrt{3})+P(-2b\sqrt{3})=2P(3b).$$We will equate coefficients of $b^{2n}$ on both sides, noting that $P$ is even. If the coefficient of $x^{2n}$ in $P$ is $a_2n$, then we have $$a_{2n}(2\cdot 3^n+12^n)=a_{2n}(2\cdot 9^n).$$Thus, or each nonnegative integer $n$, either $$a_{2n}=0$$or $$2\cdot 3^n+12^n=2\cdot 9^n.$$This is only true for $n=1$ and $n=2$, as we can test that $n=0$ doesn't work, and $12^n>2\cdot 9^n$ for $n\geq 3.$ Thus, only the $x^2$ and $x^4$ coefficients are allowed to be nonzero, thus done. remark: The main idea here is to make a substitution that, combined with the constraint of $ab+ac+bc=0$, makes it so that all terms are of the form $$kP(rb).$$This is very nice because this means that the $x^n$ coefficent of this only depends on the $x^n$ coefficient of $P$, allowing us to easily equate coefficients. By only setting "homogeneous" constraints, like $a-b=b-c$ that I used, one can prevent the input of the polynomial from having a constant shift. In fact, $$(a,b,c)=(b(1+\sqrt{3}),b,b(1-\sqrt{3}))$$is really the only substitution you need to solve this problem.

Cusofay wrote: We can check that $P(x)=mx^4+nx^2$ for real numbers $m,n$ works and we want to prove its uniqueness. Note that if $P$ and $Q$ are solutions, $aP+bQ$ is a solution for all real numbers $a,b$ so it's enough to find all solutions of the form $x^n$.$P(0,0,0)$ and $P(0,0,x)$ gives us that $n$ is even $$P(6,3,-2) \Rightarrow 3^n+5^n+(-8)^n=2 \cdot 7^n$$ If $n=0$, then this obviously doesn't hold. If $n=2$, then $9+25+64= 98 =2 \cdot 49$, which works. If $n=4$, then $81+625+4096 = 4802 = 2 \cdot 2401 $, which also works. If $n=6$, then $$3^6+5^6+8^6>8^6=(7+1)^6=7^6+\frac{6}{7} \cdot 7^6 + \frac{15}{49} \cdot 7^6 = (2+\frac{8}{49}) \cdot 7^6>2 \cdot 7^6,$$which is a contradiction. If $n \ge 8$, then $$3^n+5^n+8^n>8^n=(7+1)^n>7^n+n \cdot 7^{n-1} > 2 \cdot 7^n,$$contradiction. $$\mathbb{Q.E.D.}$$ unfortunately, the linear combination of any two solutions also being a solution does not necessarily imply that checking $x^n$ suffices for example, if hypothetically the solution set is $r(x+1)$ for real numbers $r$, 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 $x^n$ even though there does exist solutions. the issue here is that $1,x,x^2,\dots$ 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 it's ok though I wish you good luck in the future!

The answer is $f(x)=ux^4+vx^2$ where $u,v\in \mathbb R$. This polynomial indeed satisfies the problem. We get from the equation that $f(0)=0$ and $f(x)=f(-x)$ for all $x\in \mathbb R$. So we can write $$f(x)=\sum_{k=1}^n a_k x^{2k}$$where $a_n\neq 0$. Now we put $(a,b,c)=\left(x,-\frac{x}{1+x},1\right)$ into the equation and after clearing the denominator we get $$\sum_{k=1}^n a_k Q_k(x)=0\qquad(\star)$$for all $x\in \mathbb R$, where $$Q_k(x)=x^{2k}(x+2)^{2k}+(2x+1)^{2k}+(x^2-1)^{2k}-2(x^2+x+1)^{2k}.$$We can check that $Q_2(x)=Q_4(x)=0$ for all $x$. And for $k>2$ we can prove via the multinomial theorem that $\deg Q_k=4k-2$. Now we claim that $n\leq 2$, which will finish the problem. Suppose ftsoc that $n>2$ and let $\deg Q_n=m$. Since $m>\max_{1\leq k\leq n-1} \deg Q_k$, it follows that the coefficient of $x^m$ in $(\star)$ is some nonzero number times $a_n$. Thus we must have $a_n=0$, which is the desired contradiction.

Am I the only one to find this problem very easy for a P2 or infact even a P1?

Let $S$ be the set of solutions. We prove that $S = \{rx^2 + sx^4 \mid|r, s\in\mathbb R\}$ $(0, 0, 0) \implies P(0) = 0, (0, 0, c) \implies P(c)= P(-c),$ so $P$ is an even function. Claim: $S$ is a $\mathbb R$-vector space Proof: This is obvious. Claim:$P(x) = x^2, x^4$ both work. Proof: Both are just expansion Claim: $\deg P \leq 4$ Proof: Take $(6x, 3x, -2x)$ to get $P(3x) + P(5x) + P(8x) = 2P(7x).$ Hence, if a polynomial works, then we require $3^{2n} + 5^{2n} + 8^{2n} = 2\cdot7^{2n}$ where $n = \deg P.$ However, simply note that $8^{2n}\geq2\cdot7^{2n}$ for $n\geq3.$