Determine all $P(x) \in \mathbb R[x]$ such that $P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2$, $\forall x \in \mathbb R$.

Question

Determine all polynomials $P(x) \in \mathbb R[x]$ knowing that $$ P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2, \forall x \in \mathbb R\,.$$

Replacing $x$ by $-x$, we have that $$P(x^2) - x[mP(-x) + nP(x)] = [P(-x)]^2 + (m - n)x^2, \forall x \in \mathbb R$$

Subtracting the second equation from the first, $$x(m + n)[P(x) + P(-x)] = P(x)^2 - P(-x)^2, \forall x \in \mathbb R$$

$$ \iff \left[ \begin{align} P(x) + P(-x) &= 0\\ P(x) - P(-x) &= x(m + n) \end{align} \right.$$

In the case of $P(x) +P(-x)=0$, we have that $$P(x^2) + (m - n)xP(x) = [P(x)]^2 + (m - n)x^2$$

In the case of $P(x) - P(-x) = x(m + n)$, we have that $$P(x^2) + x(m + n)[P(x) -nx] = [P(x)]^2 + (m - n)x^2$$

That neither of which can I answer is not unexpected.

Answer 1

We conjecture that for odd degree $s\geq 3$ and above, the only polynomials that work are

$$P(x)=x^s+\frac{m-n}{2}x\text{ for }(m - n) (-2 + m - n)=0$$

while for even degree $s>3$ the only polynomials that work are

$$P(x)=x^s+\frac{m+n}{2}x\text{ for }(m - 3 n) (-2 + m + n)=0$$

In order to prove this, let

$$P(x)=\sum_{i=0}^s a_i x^i$$

where $a_s\neq 0$ and $s\geq 3$. Then by the equation given

$$0=P(x^2) + x (m P(x) + n P(-x))-P(x)^2-(m-n)x^2$$

$$=\sum_{i=0}^s a_i x^{2i}+\sum_{i=0}^s ma_i x^{i+1}+\sum_{i=0}^s na_i (-1)^ix^{i+1}-\sum_{i=0}^s\sum_{j=0}^sa_ia_jx^{i+j}-(m-n)x^2$$

We compute the highest coefficients first. Obviously, the highest power of $x$ is at most $2s$. This coefficient is

$$0=a_s-a_s^2\Rightarrow 0=a_s-1$$

This implies $a_s=1$. For $x^{2s-1}$, we have

$$0=-2a_{s-1}a_s=-2a_{s-1}$$

This implies $a_{s-1}=0$. For $x^{2s-2}$ we have

$$0=a_{s-1}-2a_{s-2}a_s-a_{s-1}^2=-2a_{s-2}$$

Again, this implies $a_{s-2}=0$. By induction, this pattern holds until $a_1$. It is this induction step that we use the fact that $s\geq 3$, else the pattern would break from $(m-n)x^2$. That is, we know

$$a_{s-1}=a_{s-2}=\cdots a_3=a_2=0$$

Then the polynomial is of the form

$$P(x)=x^s+ ax+b$$

Putting this into the equation gives us

$$0=(b-b^2)+x b (-2 a + m + n)+x^2(1-a) (a - m + n)+x^s(-2b)+x^{s+1}(m+n(-1)^s-2a)$$

Obviously, $b=0$. Then the equation becomes

$$0=x^2(1-a) (a - m + n)+x^{s+1}(m+n(-1)^s-2a)$$

Then

$$a=\frac{m+n(-1)^s}{2}$$

This gives us the first part of our conjecture. For the second part

$$0=(1-a) (a - m + n)=(2-m+n(-1)^s)(m+n(-1)^s-2m+2n)$$

Thus, either

$$2-m+n(-1)^s=0$$

or

$$m+n(-1)^s-2m+2n=0$$

For degree $0$, we have

$$P(x)=a$$

for some $a\in\mathbb{R}$. This implies

$$0=(a-a^2)+ax(m+n)+x^2(n-m)$$

Since this holds for all $x$, we know $a\in\{0,1\}$ and $m-n=0$. Without knowing more about $m$ and $n$, this is the best we can do. Now, let

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

for $a\neq 0$. Then

$$0=b(1-b)+xb (-2 a + m + n)+x^2(1 - a) (a - m + n)$$

This implies

$$b\in\{0,1\}$$

but other than than we can't say anything else without more restrictions on $m$ and $n$. Note that $(a,b)=(1,0)$ will work no matter what $n$ and $m$ are. For quadratics, let

$$P(x)=ax^2+bx+c$$

for $a\neq 0$. Then

$$0=c - c^2 + (-2 b c + c m + c n) x + (b - b^2 - 2 a c - m + b m + n - b n) x^2 + (-2 a b + a m + a n) x^3 + (a - a^2) x^4$$

Since $a\neq 0$, this implies $a=1$. Then

$$0=c - c^2 + (-2 b c + c m + c n) x + (b - b^2 - 2 c - m + b m + n - b n) x^2 + (-2 b + m + n) x^3$$

which implies

$$b=\frac{m+n}{2}\text{ and }c\in\{0,1\}$$

Then

$$P(x)=x^2+\frac{m+n}{2}x+c\text{ for }-2 c + \frac{1}{4} (m - 3 n) (-2 + m + n)=0\text{ with }c\in\{0,1\}$$

Having checked all degrees, let us discuss the results. First, if $m,n$ are arbitrary, then the only polynomial that works is $P(x)=x$. Second, we can list out all equations of $m$ and $n$ that give infinite solutions. We have

$$m - n=0$$

$$m - 3n=0$$

$$m+n=2$$

$$m-n=2$$

That is, if $m$ and $n$ do not satisfy one of these equations, then $P(x)$ has only finite solutions. The final equation which also gives a finite solution set is

$$-8 - 2 m + m^2 + 6 n - 2 m n - 3 n^2=0$$

which creates the quadratic

$$P(x)=x^2 + \frac{m + n}{2} x + 1$$

Answer 2

I think you got wrong in the first step.Replacing $x$ by $−x$, we have indeed $P(x^2) - x[mP(-x) + nP(x)] = [P(-x)]^2 + (m - n)x^2, \forall x \in \mathbb R$ but not $P(x^2) + x[mP(-x) + nP(x)] = [P(-x)]^2 + (m - n)x^2, \forall x \in \mathbb R$.

So after substracting we should get $x(m + n)[P(x) + P(-x)] = P(x)^2 - P(-x)^2$.

That's $P(x) + P(-x) = 0$ or $P(x) - P(-x) = x(m + n)$.

In the former case try writing $P(x) = xF(x^2)$, and the latter $P(x) = F(x^2) + \frac{m+n}{2}x$, where $F[x]\in \mathbb R[x]$

Answer 3

This is a sketch. I will polish it when I have time.

As in Zerox's answer, we can easily see that $P(x)$ is an odd polynomial (so $P(x)=x\,Q(x^2)$ for some polynomial $Q(x)$), or $P(x)=\dfrac{m+n}{2}\,x+R(x^2)$ for some polynomial $R(x)$. We deal with each case separately.

In the first case, we see that $$x^2Q(x^4)+(m-n)\,x^2\,Q(x^2)=\big(x\,Q(x^2)\big)^2+(m-n)\,x^2\,.$$ Since $Q(x)$ is a polynomial, we get $$Q(x^2)+(m-n)\,Q(x)=\big(Q(x)\big)^2+(m-n)\,.$$ That is, $$Q(x^2)+(m-n)\,Q(-x)=\big(Q(-x)\big)^2+(m-n)\,.$$ Subtracting the two equations above yields $$(m-n)\,\big(Q(x)-Q(-x)\big)=\big(Q(x)\big)^2-\big(Q(-x)\big)^2\,.$$ Thus, $$Q(x)-Q(-x)=0$$ or $$Q(x)+Q(-x)=m-n.$$ If $Q(x)$ is constant, then $Q(x)=1$ or $Q(x)=m-n$, so that $$P(x)=x$$ or $$P(x)=(m-n)x\,.$$ If $Q(x)$ is nonconstant, then we can let $Q(x)=S(x^{2^k})$, where $k$ is the largest possible nonnegative integer and $S(x)$ is a polynomial. Then, we must have $S(x)+S(-x)=m-n$. That is, $$S(x)=\dfrac{m-n}{2}+x\,S_1(x^2)$$ for some polynomial $S_1(x)$ which must satisfy $$\frac{m-n}{2}+x^2\,S_1(x^4)+(m-n)\left(\frac{m-n}{2}+x\,S_1(x^2)\right)=\left(\frac{m-n}{2}+x\,S_1(x^2)\right)^2+m-n\,.$$ Hence, $$x^2\,S_1(x^4)+\frac{m-n}{2}\left(\frac{m-n}{2}-1\right)=x^2\,\big(S_1(x^2)\big)^2\,.$$ Therefore, the constant terms of the two sides of the equation above must coincide, making $m=n$ or $m=n+2$. In the case $m=n$, we must have $S_1(x^2)=\big(S_1(x)\big)^2$. That is, $S_1(x)=x^d$ for some integer $d\geq 0$, so that $S(x)=x^{2d+1}$. This means $Q(x)=x^{2^k(2d+1)}$, so that in the case $m=n$, we have this extra solution: $$P(x)=x\,Q(x)=x^{2^k(2d+1)+1}\,,$$ where $k\geq 1$ and $d\geq 0$ are integers. In other words, $P(x)=x^t$ with $t\geq 3$ being an odd positive integer is a solution when $m=n$. In the case $m=n+2$, we must have again that $S_1(x)=x^d$ for some integer $d\geq 0$, making $S(x)=1+x^{2d+1}$. Ergo, $Q(x)=1+x^{2^k(2d+1)}$, and $$P(x)=x\,Q(x)=x+x^{2^k(2d+1)+1}=x+x^t\,,$$ where $t\geq 3$ is an odd positive integer. Thus, all solutions where $P(x)$ is an odd polynomial are given by

• $P(x)=x$,
• $P(x)=(m-n)x$,
• $P(x)=x^{2s+1}$ if $m=n$, where $s$ is a positive integer, and
• $P(x)=x+x^{2s+1}$ if $m=n+2$, where $s$ is a positive integer.

In the second case, we see that $$\frac{m+n}{2}\,x^2+R(x^4)+(m+n)\,x\,\left(\frac{m+n}{2}\,x+R(x^2)-nx\right)=\left(\frac{m+n}{2}\,x+R(x^2)\right)^2+(m-n)\,x^2\,.$$ If $r$ is the constant term of $R(x)$, then $r=r^2$ must hold. Thus, $r=0$ or $r=1$. If $r=0$, the coefficients of $x^2$ of both sides of the equations above must be the same. That is, $$\frac{m+n}{2}+(m+n)\,\left(\frac{m+n}{2}-n\right)=\left(\frac{m+n}{2}\right)^2+(m-n)\,.$$ That means $m=-n+2$ or $m=3n$. For these values of $m$, $R(x^2)=\big(R(x)\big)^2$, so that $R(x)=x^s$ for some positive integer $s$. That is, $$P(x)=\frac{m+n}{2}\,x+x^{2s}$$ for some positive integer $s$ if $m=-n+2$ or $m=3n$. Let now assume that $r=1$. Then, we write $R(x)=1+x\,R_1(x)$. That is, $$\frac{m+n}{2}\,x^2+1+x^4\,R_1(x^4)+(m+n)\,x\,\left(\frac{m+n}{2}\,x+1+x^2\,R_1(x^2)-nx\right)=\left(\frac{m+n}{2}\,x+1+x^2\,R_1(x^2)\right)^2+(m-n)\,x^2\,.$$ That is, $$x\,R_1(x^2)+\frac{1}{4}(m-3n)(m+n-2)=x\,\left(R_1(x)\right)^2+2\,R_1(x)\,.$$ It follows that $R_1(x)+R_1(-x)$ divides $\dfrac{1}{2}(m-3n)(m+n-2)$. We conclude that $R_1(x)+R_1(-x)$ must be constant. That is, $$R_1(x)=a+x\,U(x^2)$$ for some polynomial $U(x)$ and for some constant $a$. Clearly, $a=\dfrac{1}{8}(m-3n)(m+n-2)$, and $$a+x^2\,U(x^4)=\big(a+x\,U(x^2)\big)^2+2\,U(x^2)\,.$$ This shows that $2ax\,U(x^2)$ is an even polynomial. Therefore, either $a=0$ or $U(x)=0$. If $a=0$, then $$x\,U(x^2)=x\,\big(U(x)\big)^2+2\,U(x)\,.$$ It follows easily that $U(x)$ must equal $0$ by considering the coefficient of the nonzero lowest term of $U(x)$ if exists. Thus, in any case, $R_1(x)=a=\dfrac{1}{8}(m-3n)(m+n-2)$. However, we also need $a^2=a$, so that $a=0$ or $a=1$, making $m=3n$, $m=-n+2$, or $m=n+1\pm\sqrt{4n^2-4n+9}$ (which only has the following integer solutions $(m,n)$: $(-2,0)$, $(4,0)$, $(-1,1)$, and $(5,1)$). That is, $$P(x)=1+\frac{m+n}{2}\,x+\frac{(m-3n)(m+n-2)}{8}\,x^2$$ in this case. In conclusion, the solutions in the case where $P(x)=\dfrac{m+n}{2}\,x+R(x^2)$ are given by

• $P(x)=\dfrac{m+n}{2}\,x+x^{2s}$ if $m=3n$ (corresponding to $P(x)=2n\,x+x^{2s}$) or $m=-n+2$ (corresponding to $P(x)=x+x^{2s}$), where $s$ is a nonnegative integer, and
• $P(x)=1+\dfrac{m+n}{2}\,x+x^2$ if $m=n+1\pm\sqrt{4n^2-4n+9}$ (which only has the following integer solutions $(m,n)$: $(-2,0)$ corresponding to $P(x)=1-x+x^2$, $(4,0)$ corresponding to $P(x)=1+2x+x^2$, $(-1,1)$ corresponding to $P(x)=1+x^2$, and $(5,1)$ corresponding to $P(x)=1+3x+x^2$).

Anyway, if we parametrize $(m,n)$ such that $m=n+1\pm\sqrt{4n^2-4n+9}$ via $$(m,n)=\left(\dfrac{3q^2+6q+8}{4q},\dfrac{q^2+2q-8}{4q}\right)\,,$$ where $q$ is a nonzero real number. That is, $$P(x)=1+\left(1+\frac{q}{2}\right)\,x+x^2$$ is a solution.