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Determine all polynomials $P(x)$ with real coefficients satisfying the following conditions:

  • $P(2020) = 2019$
  • $(P(x) + 1)^2 = P(x^2 + 1)$ for all $x\in\mathbb{R}$

I wrote down $p(x$) as $a_nx^n +\ldots+a_0$ and then started applying substituting $2020$ and tried to get awkward results. I know this is absolutely wrong. Kindly provide a rigorous answer

cmk
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  • A simple solution that meets condition 1) is $p(x)=x-1$. As it turns out, this also satisfies the second condition. – PierreCarre Dec 08 '20 at 19:54
  • @PierreCarre It looks like this is the unique solution – Raffaele Dec 08 '20 at 19:59
  • @Raffaele It seems that way but I did not do any proof... It just seemed to tempting to try this solution given the condition $p(2020)=2019$. – PierreCarre Dec 08 '20 at 20:00
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    Set $Q(x)=P(x)+1$ and use https://math.stackexchange.com/questions/271337 – user26857 Dec 08 '20 at 20:01
  • Okay... So what is the RHs coming to... Can you tell a little bit... The LHS is (Q(x))^2 –  Dec 08 '20 at 20:08
  • This looks like a contest problem. Where is it from? If it's from an old contest, there is probably a thread on AOPS, and if it's from a currently ongoing one, you shouldn't be asking about it. – Aaron Dec 08 '20 at 21:03
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    It's a variant of problem from Austrian contest 2017, anyway the point is you can determine infinitely many points of the polynomial by the second condition, so the polynomial is unique (hence $P(x)=x-1$ must be the only one). – Sil Dec 08 '20 at 22:30

2 Answers2

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I define $(x_n)$ by: $\ x_0=2020 \ $ and $\ \forall n \in \mathbb N \ , \ x_{n+1}=x_n^2 +1$.

We can show by induction that: $\ \forall n \in \mathbb N \ , P(x_n)=x_n-1$

$P(x)-(x-1)$ is a polynomial and has an infinity of roots.

And we can conclude that $P(x)=x-1$.

perroquet
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►Degree $0:\hspace0.5cm P(x)=c$. One must have $(c+1)^2=c\iff c^2+c+1=0$ so $c$ is not real and there are not solutions.

►Degree $1:\hspace0.5cm Px)=ax+b$. One must have $(ax+b+1)^2=a(x^2+1)+b$ for all $x$. Then $$a^2-a=0,\hspace0.5cm2a(b+1)=0,\hspace0.5cm(b+1)^2-a-b=0$$ whose only solution with $a\ne0$ is $(a,b)=(1,-1)$ so $P(x)=x-1$ could be a solution. Since $P(2020)=2019$ we have a solution of the first degree $\boxed{P(x)=x-1}$.

►Degree $2:\hspace0.5cm P(x)=x^2+bx+c$ (for degree greater than or equal to two, clearly the leading coefficient should be equal to $1$ because $a_n^2 = a_n$). One must have $(x^2+bx+c+1)^2=(x^2+1)+b(x^2+1)+c$ for all $x$. Then $$x^4+b^2x^2+(c+1)^2+2bx^3+2(c+1)x^2+2b(c+1)x=x^4+2x^2+1+bx^2+b+c$$ so $b=0$ and $2(c+1)=2$ then $c=-1$. Thus $P(x)=x^2-1$ but then $P(2020)\ne2019$ so there are not solutions for quadratic polynomials.

►Degree $n\gt2:\hspace0.5cm P(x)=x^n+Q(x)$ where $Q(x)$ is of degree at most $n-1$. One must have $$(x^n+Q(x)+1)^2=(x^2+1)^n+Q(x^2+1)\text { for all x}.$$ $$\left(x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_2x^2+a_1x+(a_0+1)\right)^2=(x^2+1)^n+\cdots+a_1(x^2+1)+a_0$$ Since all powers in RHS are even, all coefficient of all odd power in LHS must be null. We get $$a_1(a_0+1)=0\\a_3(a_0+1)+a_1a_2=0\\a_5(a_0+1)+a_1a_4+a_2a_3=0\\a_7(a_0+1)+a_1a_6+a_2a_5+a_3a_4=0\\a_9(a_0+1)+a_1a_8+a_2a_7+a_3a_6+a_4a_5=0\\.......................\\.......................\\a_{2n-1}(a_0+1)+a_1a_{2n-2}+a_2a_{2n-3}+\cdots+a_{n-1}a_{n}=0 \text{ where $a_n=1$}$$ from which it is easily deduced $a_1^2a_2=0,\space a_1a_2^2a_3=0,\space a_1a_2a_3^2a_4=0,......$

Do these necessary conditions lead to $P(x)=x^n-1$ in which case $P(2020)\ne{2019}$?

We see the cubic case separately and leave the general case which offers some difficulty in handling the coefficients for whoever wants to solve it, mainly for those who have suitable electronic calculators. It is not unlikely that there is any artifice to resolve this problem in an easier way.

On the other hand,it seems very likely that the only solution is $P(x)= x-1$.

►Degree $3:\hspace0.5cm P(x)=x^3+bx^2+cx+d$. $$(x^3+bx^2+cx+d+1)^2=(x^2+1)^3+b(x^2+1)^2+c(x^2+1)+d$$ Equating to $0$ the coefficients of $x$ and $x^3$ on the left side and the coefficients of $x^2$ on both sides, we have the system $$\begin{cases}d+1+bc=0\\c(d+1)=0\\c^2+2bd-c=3\end{cases}$$ whose solutions are $$(a,b,c)=(-1.5,0,-1),(0,\frac{1+\sqrt{13}}{2},-1),(0,\frac{1-\sqrt{13}}{2},-1)$$ and it is verified that none of the three resulting cubic polynomials satisfies $P(2020)=2019$. Thus there is no solution for degree $3$.

Piquito
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