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Find polynomial $f(x)$ with real coefficients that satisfied:

$$x^2f(x) + 2 = f(x^2) + 2x^3$$

I find that $\deg f$ can be $1$ and $2$.

$$\deg f = n $$

$$2+n=\max(2n,3)$$

  1. First case $2=n$

  2. Second case $n=1$


But what next?

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    $\begingroup$ @SchrodingersCat : If you're going to do an edit whose only content is to fix some blatant MathJax solecisms, why do you leave equally blatant MathJax solecisms intact, such as writing $deg f$ rather than $\deg f$ (with \deg) or f(x) instead of $f(x)$, or $max(2n,3)$ instead of $\max(2n,3)$? ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 20 '16 at 17:30
  • $\begingroup$ It shouldn't be too hard to check if :$$f(x)=ax^2+bx+c$$ works (where $a$ can be $0$ when $\deg f=1$ ) . In this way you can find $a,b,c$ . Are you looking for an elegant way ? $\endgroup$ – user252450 Jan 20 '16 at 17:32
  • $\begingroup$ @MichaelHardy I just corrected the equation. I am not a MathJax corrector. I only correct parts of the question when it cannot be understood well. $\endgroup$ – SchrodingersCat Jan 20 '16 at 17:32
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Your polynomial has the form $f(x)=ax^2+bx+c$ with $a,b,c\in \mathbb{R}$.

Thus your equation gives $$ax^4+bx^3+cx^2+2=ax^4+bx^2+c+2x^3$$and comparing the terms of each degree on both sides yields $b=c=2$, and $a$ can be any real number.

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If $n=0$, the degree on the left is 2 and on the right is 3, so there are no solutions.

If $n = 1$, let $f(x) = ax + b$. Solve $$x^2(ax+b) = ax^2 + b + 2x^3$$ for a and b.

If $n = 2$, let $f(x) = ax^2+bx+c$. Plug in and solve for $a$, $b$, and $c$.

If $n >= 3$, $max(2n,3) = 2n$. Since $n > = 3 => n+n >= 3+n>2+n$, the degree on the right is larger than the degree on the left. So there are no further solutions.

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