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This problem is found in a past test.

Let $n$ be a positive integer. Find all polynomials P with real coefficients such that

$P(x^2 + x-n^2) = P(x)^2 +P(x)$

for all real numbers x.

I think that we should be comparing coefficients of certain terms. If we start by noting the degrees of the polynomials involved,

If P is a polynomial of degree d , the left-hand side $P(x^2 + x - n^2) $ has degree 2d . The right-hand side $ P(x)^2 + P(x) $ also has degree 2d.

If P is a constant polynomial c , then $c = c^2 + c $, leading to c = 0, so we have P(x) = 0.

If P(x) is linear and $ P(x) = ax + b $. Substituting into the equation, the coefficients lead to a contradiction unless n = 0 . If we have $ P(x) = ax^2 + bx + c $, substitution also rules this out.

Are there better ways of proceeding and coming up with a contradiction that may lead to the answer being $P(x)=0$ or are there other answers obtained in higher powers?

Thanks in advance.

John O'neil
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  • No, it is not. I found it from a past test, not a current one. I think I accidentally typed the wrong competition and mistook it for this question: https://artofproblemsolving.com/community/c6h1141745p5366265 , I will now edit the question accordingly @user1551 – John O'neil Feb 18 '25 at 10:55

2 Answers2

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From the equation we see that $\{-n,n\}$ are (distinct, since $n > 0$) roots of $P(x)$. Then $P(x) = (x^{2}-n^{2})Q(x)$. Thus, $$(x^{2} + x - n^{2} - n)(x^{2}+x-n^{2}+n)Q(x^{2}+x-n^{2}) = \\ = (x^{2}-n^{2})^{2}Q^{2}(x) + (x^{2}-n^{2})Q(x) \Leftrightarrow (x^{2}-n^{2})(x+n+1)(x-n+1)Q(x^{2}+x-n^{2}) =\\ = (x^{2}-n^{2})^{2}Q^{2}(x) + (x^{2}-n^{2})Q(x),$$ thus, $$(x+n+1)(x-n+1)Q(x) = (x^{2}-n^{2})Q^{2}(x) + Q(x)$$ Plugging $x = n$ yields $$(2n+1)Q(n) = Q(n),$$ so $Q(n) = 0$. Plugging $x=-n$ yields $$(-2n+1)Q(-n) = Q(-n),$$ so $Q(-n) = 0$. It follows that $$P(x) = (x^{2}-n^{2})^{2}R(x),$$ doing the exact same thing you get that $x^{2}-n^{2}\mid R(x)$ and so on, eventually you come to a contradiction, since the degree is finite. Let's prove that by induction.

Claim: if $P_{k}(x) = (x^{2}-n^{2})^{k}Q_{k}(x),\;k\in\mathbb{N}_{0}$ satisfies the equation, then $Q_{k}(x) = (x^{2}-n^{2})Q_{k+1}(x)$ for some polynomial $Q_{k+1}(x)$.

Proof: the base case $k=0$ was covered above. Suppose that the claim is true for some $k\in\mathbb{N}$. Then $$P_{k}(x^{2}+x-n^{2}) = (x^{2}+x-n^{2}-n)^{k}(x^{2}+x-n^{2}+n)^{k}Q_{k}(x^{2}+x-n^{2}) = \\ = (x^{2}-n^{2})^{k}(x-n+1)^{k}(x+n+1)^{k}Q_{k}(x^{2}+x-n^{2}),$$ and $$(P_{k}(x))^{2} + P_{k}(x) = (x^{2}-n^{2})^{2k}(Q_{k}(x))^{2} + (x^{2}-n^{2})^{k}Q_{k}(x),$$ so $$(x-n+1)^{k}(x+n+1)^{k}Q_{k}(x^{2}+x-n^{2}) = (x^{2}-n^{2})^{k}(Q_{k}(x))^{2} + Q_{k}(x)$$ Now setting $x = n$, we obtain $$(2n+1)^{k}Q_{k}(n) = Q_{k}(n)\implies Q_{k}(n)=0,$$ and similarly, the substitution $x = -n$ gives $$(-2n+1)^{k}Q_{k}(n) = Q_{k}(n)\implies Q_{k}(-n)=0,$$ thus, $$x^{2}-n^{2}\mid Q_{k}(x)\;\;\;\blacksquare$$

chirico
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  • To me, it's not immediately obvious that we have the infinite descent of $(x^2 -n^2) ^ k \mid P(x)$, in part because the coefficients seem to keep on changing. As such, I would encourage you to lay out the structure of the equations satisfied by $Q(x) = P_1(x), R(x) = P_2(x)$, etc. – Calvin Lin Feb 18 '25 at 17:05
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    IE The details are: Proof by induction. Suppose $ P(x) = (x - n)^k(x+n)^k P_k (x)$, then $P(x^2 + x - n^2) = (x^2 +x - n^2 - n)^k(x^2 + x - n^2 + n)^k P_k(x^2+x - n^2) $ and $P(x)^2 + P(x) = (x-n)^k(x+n)^k P_k(x) [ (x-n)^k(x+n)^k P_k(x) + 1 ] $. This gives us $(x+n+1)^k(x-n+1)^kP_k(x^2 + x - n^2) = P_k(x) [ (x-n)^k(x+n)^k P_k(x) + 1 ]$. Substituting in $x = n$, we get $(2n+1)^k P_k(n) = P_k(n) $ and thus $P_k(n) = 0 $. Likewise for $ x = - n$. So $ P_k(x) = (x-n)(x+n)P_{k+1} (x)$. – Calvin Lin Feb 18 '25 at 17:15
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(This is basically Calvin Lin’s and chirico’s idea, but I try to avoid infinite descent.)

Let $f(x)=x^2+x-n^2$. The given condition can be rephrased as $$ P(f(x)) = P(x)^2 + P(x).\tag{1} $$ Since $n$ and $-n$ are fixed points of $f$, we obtain $P(\pm n)=0$. Therefore $$ P(x)=(x-n)^r\ (x+n)^s\ Q(x)\tag{2} $$ for some positive integers $r,s$ and some polynomial $Q$ such that either $Q=0$ (in which case $P$ is the zero polynomial) or both $Q(n)$ and $Q(-n)$ are nonzero. Hence $(1)$ and $(2)$ give \begin{align*} (f(x)-n)^r\ (f(x)+n)^s\ Q(f(x)) &= P(x)\left(P(x)+1\right),\\ \left[(x-n)\ (x+1+n)\right]^{\,r}\ \left[(x+n)\ (x+1-n)\right]^{\,s}\ Q(f(x)) &= P(x)\left(P(x)+1\right),\\ (x+1+n)^r\ (x+1-n)^s\ Q(f(x)) &= Q(x)\left(P(x)+1\right). \end{align*} Putting $x=n$ into last equality, we get $(2n+1)^r\ Q(n) = Q(n)$. Hence $Q(n)=0$ and by the definition of $Q$, we must have $Q=0$. Consequently, $P=0$.

user1551
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