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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 '18 at 3:58
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    $\begingroup$ @abc... Why does that matter? $\endgroup$ – Jaideep Khare Jun 14 '18 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 '18 at 4:09
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    $\begingroup$ @HarshKatara I am sure that you are missing some assumptions on $f$. $\endgroup$ – David Jun 14 '18 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 $n\in\Bbb{Z}$. $\endgroup$ – Jyrki Lahtonen Jun 14 '18 at 5:56
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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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    $\begingroup$ Thank you very much! $\endgroup$ – Peerakorn Trechakachorn Apr 26 '20 at 6:54
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    $\begingroup$ Brilliant solution. $\endgroup$ – Boka Peer Apr 26 '20 at 7:25
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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 '18 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. $\endgroup$ – Mohammad Riazi-Kermani Jun 14 '18 at 4:21
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    $\begingroup$ Why could you assume that $f$ is linear? $\endgroup$ – Ariel Serranoni Jun 14 '18 at 4:22
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    $\begingroup$ That was the easiest case to consider. The degrees at both sides match for linear functions. $\endgroup$ – Mohammad Riazi-Kermani Jun 14 '18 at 4:24
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    $\begingroup$ Isn't there a loss of generality? $\endgroup$ – Harsh Katara Jun 14 '18 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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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. $\endgroup$ – Ariel Serranoni Jun 14 '18 at 7:02
  • $\begingroup$ Yes, if we substitute the '2' exponent by '3'. $\endgroup$ – Ariel Serranoni Jun 14 '18 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 $\endgroup$ – MaximusFastidiousIrreverence Jun 14 '18 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 '18 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 $\endgroup$ – MaximusFastidiousIrreverence Jun 15 '18 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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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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  • $\begingroup$ Thanks a lot! I appreciate your hint! $\endgroup$ – Peerakorn Trechakachorn Apr 26 '20 at 6:55
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    $\begingroup$ How do you use this hint? Can you please expand this into a sketch of a solution? $\endgroup$ – Batominovski Apr 26 '20 at 6:59
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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. $\endgroup$ – Batominovski Apr 26 '20 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. $\endgroup$ – Batominovski Apr 26 '20 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. $\endgroup$ – Batominovski Apr 26 '20 at 8:20

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