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Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.

Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x$ then find $f(2016)$. (The question is now deleted.) The same problem with finding $2011$ (instead of $2016$) appeared in 2011 Singapore Mathematical Olympiad as problem 17 (Wayback Machine).

I’ve tried put $x=0,1$ and got \begin{align*} f(0)+2f(2)&=0\\ f(2)+2f(0)&=-6 \end{align*} which gives me $f(0)=-4$, $f(2)=2$.

Similarly, if we notice that $x^2+x=x^2-3x+2$ holds for $x=\frac12$, we can find the value at the point $\frac34=\left(\frac12\right)^2+\frac12$.

But the above doesn’t seem to help for other values.

Thank you very much for helping.

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  • $\begingroup$ What's the domain and the codomain of the function? $\endgroup$
    – abc...
    Jun 14, 2018 at 3:58
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    $\begingroup$ @abc... Why does that matter? $\endgroup$ Jun 14, 2018 at 4:00
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    $\begingroup$ $x^2+x=x(x+1),x^2-3x+2=(x-1)(x-2)$... $\endgroup$
    – abiessu
    Jun 14, 2018 at 4:09
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    $\begingroup$ @HarshKatara I am sure that you are missing some assumptions on $f$. $\endgroup$
    – David
    Jun 14, 2018 at 4:40
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    $\begingroup$ Puttiing together @abiessu's observation and Mohammad's calculation tells us that if we define $$g(x)=f(x^2+x)-\big[3(x^2+x)-4\big]$$ then the given relation can be rewritten to read $$g(x)+2g(x-2)=0.$$ We also have the obvious symmetry $g(x)=g(1-x)$. These together reproduce Dan's result rewritten in the form $g(1/2)=0$. All this implies that $g(2n+1/2)=0$ for all $n\in\Bbb{Z}$. $\endgroup$ Jun 14, 2018 at 5:56

8 Answers 8

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Replace $x$ by $1-x$ and then you can see how the equation transforms (I'll let you see it yourself). Then you solve the equations. Tell me if you need more help.

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First (observation):

Note that we can determine $f(0), f(2)$ easily: $$ x=0 \qquad \rightarrow \qquad f(0)+2f(2)=0;\\ x=1 \qquad \rightarrow \qquad f(2)+2f(0)=-6; $$ so $$ f(0)=-4,\quad f(2)=2. $$

Same way we can determine $f(6), f(20)$ (substituting $x=-3, x=4$).
Same way we can determine $f(56), f(30)$ (substituting $x=-6, x=7$).
...

Second (solution):

Let's focus on $x=-a, x=a+1$, where $a\in\mathbb{R}$: $$ x=-a \qquad \rightarrow \qquad f(a^2-a)+2f(a^2+3a+2) = 9a^2+15a; \\ x=a+1 \qquad \rightarrow \qquad f(a^2+3a+2)+2f(a^2-a) = 9a^2+3a-6; $$

so (when denote $A=f(a^2-a)$, $B=f(a^2+3a+2)$): $$ \left\{ \begin{array}{l}A+2B = 9a^2+15a; \\ B+2A = 9a^2+3a-6;\end{array} \right.$$ $$ \left\{ \begin{array}{l}B+A = 6a^2+6a-2;\\ B-A = 12a+6;\end{array} \right. $$ and $$ \left\{ \begin{array}{l}f(a^2-a) = A = 3a^2-3a-4; \\ f(a^2+3a+2) = B = 3a^2+9a+2. \end{array}\right.\tag{1} $$

From $(1)$ we conclude that for each $z$ which can be written in the form $$ z = a^2-a, \qquad a \in\mathbb{R} \tag{2} $$ (in fact, for $z\ge -\frac{1}{4}$) we have $$ f(z) = 3z-4. $$ Therefore $f(z)$ is linear function for $z\ge -\frac{1}{4}$.

Since $z=2016$ admits representation $(2)$, then $f(2016)=3\cdot 2016-4 = 6044.$

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First, we solve $x^2 + x = 2016$ and (separately) $x^2 - 3x + 2 = 2016$ and write down the solutions. Then observe that, luckily,

When $x = \dfrac{-1 - \sqrt{8065}}{2}$:

  1. $x^2 + x = 2016$
  2. $x^2 - 3x + 2 = 2020 + 2\sqrt{8065} = a$ (say)
  3. $9x^2 - 15x = 18156 + 12\sqrt{8065}$

$$f(2016) + 2f(a) = 18156 + 12\sqrt{8065}$$

When $x = \dfrac{3 + \sqrt{8065}}{2}$:

  1. $x^2 + x = a$
  2. $x^2 - 3x + 2 = 2016$
  3. $9x^2 - 15x = 18144 + 6\sqrt{8065}$.

$$f(a) + 2f(2016) = 18144 + 6\sqrt{8065}$$

From the two equations, $$4f(2016) - f(2016) = 2(18144) - 18156$$

$$\boxed{f(2016) = 6044}$$

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Consider a linear function $ f(x)=ax+b$

$$ f(x^2+x) = ax^2+ax+b$$

$$ f(x^2-3x+2)= ax^2-3ax +2a+b$$

$$ f(x^2+x)+2f(x^2-3x+2)=3ax^2-5ax +4a+3b = 9x^2 -15x$$

$$a=3, b=-4$$ $$ f(x) = 3x-4$$

$$f(2016)=6044$$

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    $\begingroup$ Is this the unique solution? $\endgroup$
    – clark
    Jun 14, 2018 at 4:18
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    $\begingroup$ This is a solution where the function is assumed to be linear. I don't know if this is the only solution or not. $\endgroup$ Jun 14, 2018 at 4:21
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    $\begingroup$ Why could you assume that $f$ is linear? $\endgroup$ Jun 14, 2018 at 4:22
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    $\begingroup$ That was the easiest case to consider. The degrees at both sides match for linear functions. $\endgroup$ Jun 14, 2018 at 4:24
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    $\begingroup$ Isn't there a loss of generality? $\endgroup$ Jun 14, 2018 at 4:25
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We can calculate this also more generally, to get the function Mohammad proposed. We have: $x^2+x=a \Rightarrow x_{1,2}=-\frac{1}{2}\pm\frac{\sqrt{4a+1}}{2}$ and $x^2-3x+2=a \Rightarrow x_{3,4}=\frac{3}{2}\pm\frac{\sqrt{4a+1}}{2}$.

Now let's use the fact that $x_{1,2}+x_{4,3}=1$ and notice the swap of the index for the second term.

Plugging $x=x_{1,2}$ into the equation, we get:

$$ f(a)+2f(x_{4,3}^2-2x_{4,3}+1-3+3x_{4,3}+2)=9x_{1,2}^2-15x_{1,2} $$ $$ f(a)+2f(a-2+4x_{4,3})=9x_{1,2}^2-15x_{1,2} $$

and by plugging $x=x_{4,3}$ into the equation, we get: $$ f(x_{1,2}^2-2x_{1,2}+1+1-x_{1,2})+2f(a)=9x_{4,3}^2-15x_{4,3} $$ $$ f(a+2-4x_{1,2})+2f(a)=9x_{4,3}^2-15x_{4,3} $$

We can easily see that $a+2-4x_{1,2}=a-2+4x_{4,3}=t$ and we have to solve the system of equations:

$$ f(a)+2f(t)=9x_{1,2}^2-15x_{1,2} $$ $$ f(t)+2f(a)=9x_{4,3}^2-15x_{4,3} $$

Summing them up gives us: $$ 3(f(a)+f(t))=9(x_{1,2}^2+x_{4,3}^2)-15(x_{1,2}+x_{4,3})=9(1-2x_{1,2}x_{4,3})-15 $$ $$ f(a)+f(t)=3(1-2x_{1,2}(1-x_{1,2}))-5=6x_{1,2}^2-6x_{1,2}-2 $$

Subtracting them gives us: $$ f(a)-f(t)=9(x_{4,3}^2-x_{1,2}^2)-15(x_{4,3}-x_{1,2})=9(x_{4,3}-x_{1,2})(x_{4,3}+x_{1,2})-15(x_{4,3}-x_{1,2}) $$ $$ f(a)-f(t)=9(x_{4,3}-x_{1,2})-15(x_{4,3}-x_{1,2})=6(x_{1,2}-x_{4,3})=12x_{1,2}-6 $$

If we now add these two equations we get the solution: $$ 2f(a)=(6x_{1,2}^2-6x_{1,2}-2)+(12x_{1,2}-6)=(6x_{1,2}^2+6x_{1,2})-8=6a-8 $$

And finally: $f(a)=3a-4$

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Hint.

As $x^2-3x+2 = (x-2)^2+(x-2)$ calling $F(x) = f(x^2+x)$ we have

$$ F(x)+2F(x-2)=3x(3x-5) $$

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    $\begingroup$ How do you use this hint? Can you please expand this into a sketch of a solution? $\endgroup$ Apr 26, 2020 at 6:59
  • $\begingroup$ A particular solution is $$F(x) = 3x^2+3x-4$$ with null constants. Hence $f(2016) = 6044$ Don't forget. I left it as a hint. $\endgroup$
    – Cesareo
    Apr 26, 2020 at 7:40
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    $\begingroup$ That is not how it works in this situation. For your functional equation in $F$, the best you can say is $$F(x)=3x^2+3x-4+T\Biggl(\frac{x}{2}-\left\lfloor\frac{x}{2}\right\rfloor\Biggr)\cdot (-2)^{\left\lfloor \frac{x}{2}\right\rfloor}$$ for some function $T:[0,1)\to \mathbb{R}$. How exactly do you show that $T=0$? If you give a hint, please make sure that the hint is viable. $\endgroup$ Apr 26, 2020 at 7:46
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    $\begingroup$ Yes, a particular solution is $F(x)=3x^2+3x-4$. However, it is not the only solution to your functional equation in $F$. In order to conclude $f(2016)=6044$, you have to eliminate other solutions $F$ (with $T\neq 0$). Furthermore, $T$ is definitely not necessarily constant (therefore, what you say here in the quote "the periodic term which can be considered a constant" is incorrect). Why can we ignore other solutions? That is the question I am asking you. This is very nontrivial from your hint. $\endgroup$ Apr 26, 2020 at 8:05
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    $\begingroup$ I think this discussion goes nowhere. You are not going to explain in better detail how to eliminate other possible functions $F$ from the solutions of your functional equation in $F$. Because your functional equation in $F$ is not equivalent to the OP's functional equation in $f$, there has to be a process of elimination. You are not showing this, and refusing to do so. This is by far the most crucial part of the proof if we are to use your hint. $\endgroup$ Apr 26, 2020 at 8:20
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Assuming $f $ is a polynomial, consider the degree picture:

If $ \deg [f (x)]=n $, then $\deg [f (ax^2+bx+c)]=2n$, and on the RHS we have $\deg [9x^2+15x]=2$

So in solving $2n=2$, we have that the degree of $f=1$......This shows that you can assume $f $ takes the form

$$f (x)=ax+b $$

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  • $\begingroup$ Nice! But I think this shows that assuming that $f$ is linear does not always work. For example, if we had $9x^3+15x$ on the RHS. Would there even exist a solution? I don't think so. $\endgroup$ Jun 14, 2018 at 7:02
  • $\begingroup$ Yes, if we substitute the '2' exponent by '3'. $\endgroup$ Jun 14, 2018 at 7:05
  • $\begingroup$ @ArielSerranoni Perhaps If $n=2$ and you made the leading coefficients cancel or something like that. This method is only good as far as you can play the "match the coefficients" game, but with increasing degree you get more variables to play with $\endgroup$ Jun 14, 2018 at 7:11
  • $\begingroup$ @AmateurMathPirate I have downvoted this answer for two reasons: (1) this answer is not self-contained. It piggy-backs on another answer, and doesn't fully answer the question by itself. It might be better to comment on Mohammad Riazi-Kermani's answer, and suggest that he edit this material into his answer. (2) As has been pointed out in several other answers, there are hidden assumptions about the nature of $f$ that are being made throughout (i.e. that it is linear, or that it is a polynomial). You note that this has been assumed, but don't address why we can assume it. $\endgroup$
    – Xander Henderson
    Jun 14, 2018 at 22:45
  • $\begingroup$ @XanderHenderson I thought at the time something was to be gained by justifying how Mohammad supposed it was a line. And i have no idea how to proceed assuming $f $ is not a polynomial $\endgroup$ Jun 15, 2018 at 4:40
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Denote: $x^2+x=a$. Then: $$f(a)+2f(a-4x+2)=9a-24x.$$ Plug $x=\frac12$ to get: $$f(a)+2f(a)=9a-12 \Rightarrow f(a)=3a-4.$$ Hence: $$f(2016)=3\cdot 2016-4=6044.$$

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