What's the necessary and sufficient condition for a real sequence to be written as the self-convolution of another real sequence?

Question

Definition For a sequence $a_0,a_1,\cdots,a_n$, the corresponding self-convolution is another sequence $\displaystyle b_m=\sum\limits_{i+j=m}a_ia_j$ where $0\leq m\leq 2n$.

Calculating the self-convolution of a known sequence is relatively easy. My question is that if we now have an already known REAL sequence $b_0,b_1,\cdots,b_{2n}$, under what conditions can we conclude that there exists another REAL sequence $a_0,a_1,\cdots,a_n$ such that $\forall 0\leq m\leq 2n$, there is $\displaystyle b_m=\sum\limits_{i+j=m}a_ia_j$?

This problem comes from the following question. I want to find two non-zero polynomials $f(x)$ and $g(x)$ with real coefficients that satisfies the following identity $$(x^2+A)f(x)^2+1=g(x)^2,$$ where $A$ is an arbitrary positive number. What is already known is that there's the identity $$(x^2+A)\bigg(\frac{2x}{A}\bigg)^2+1\equiv \bigg(\frac{2x^2}{A}+1\bigg)^2,$$ which means one solution can be $\displaystyle f(x)=±\frac{2x}{A},\;g(x)=±\bigg(\frac{2x^2}{A}+1\bigg)$. However, I want to know that whether there exist solutions when $f(x)$ and $g(x)$ have higher degrees, i.e., whether there exists two real sequences $p_0,p_1,\cdots,p_n$ and $q_0,q_1,\cdots,q_{n+1}$ where $n\geq 2$ such that $$(x^2+A)\bigg(\sum\limits_{i=0}^np_ix^i\bigg)^2+1\equiv \bigg(\sum\limits_{i=0}^{n+1}q_ix^i\bigg)^2.$$

Now if we let $\displaystyle P_m=\sum\limits_{i+j=m}p_ip_j,\;Q_m=\sum\limits_{i+j=m}q_iq_j$, and if we compare the coefficients of $1,x,x^2,x^3,\cdots,x^{2n}$ of both sides, we immediately have the following relationship \begin{aligned}P_{2n}&=Q_{2n+2},\\P_{2n-1}&=Q_{2n+1},\\P_{2n-2}+AP_{2n}&=Q_{2n},\\ &\;\;\vdots\\AP_1&=Q_1,\\AP_0+1&=Q_0.\end{aligned} As we can see, both sides are strongly related to self-convolution, and then the problem becomes "Can the sequence $P_{2n},P_{2n-1},P_{2n-2}+AP_{2n},\cdots,AP_1,AP_0+1$ be written as the self-convolution of another sequence?"

I've been struggled with this problem for a long time, and only a few things about this problem is known.

First, $n$ cannot be even. This is because when $n$ is even, $f(x)$ is of an even number degree and $g(x)$ is of an odd number degree, which implies $g(x)=0$ has at least one real solution. Then $g(x)^2=0$ has at least one real solution. However, $(x^2+A)f(x)^2+1=0$ does not have a real solution. So we cannot find $f(x)$ and $g(x)$ such that $(x^2+A)f(x)^2+1\equiv g(x)^2$.

Second, the strategies of solving some similar cases are known. One similar case is that it's already known that there does not exist polynomials $f(x)$ and $g(x)$ with real coefficients such that $(x^2+A)f(x)^2\equiv g(x)^2$. This is because $x=\sqrt{A}i$ is obviously a solution of $(x^2+A)f(x)^2=0$ with an odd number multiplicity. However, for the equation $g(x)^2=0$, either $x=\sqrt{A}i$ is not its solution, or $x=\sqrt{A}i$ is a solution with an even number multiplicity. Another similar case is that there does not exist polynomials $f(x)$ and $g(x)$ with real coefficients such that $x^2f(x)^2+1\equiv g(x)^2$. This is because $1=(g(x)+xf(x))(g(x)-xf(x))$, which implies that both $g(x)+xf(x)$ and $g(x)-xf(x)$ has a degree of $0$. So we can let $g(x)+xf(x)=C$ and $\displaystyle g(x)-xf(x)=\frac{1}{C}$. Then $\displaystyle g(x)=\frac{1}{2}\bigg(C+\frac{1}{C}\bigg)$ and $\displaystyle xf(x)=\frac{1}{2}\bigg(C-\frac{1}{C}\bigg)$, which is impossible.

However, I don't know whether there're other real polynomial solutions for $(x^2+A)f(x)^2+1\equiv g(x)^2$. I think starting with self-convolutions may be helpful for this problem, but it seems hard to deal with self-convolutions. All I know about self convolutions is that there's the identity $$\sum\limits_{i=0}^{2n}b_ix^i=\bigg(\sum\limits_{i=0}^na_ix^i\bigg)^2,$$ which doesn't seem useful to this problem. Please help if you know something about self-convolution, or you can also use some other methods to prove or disprove that there does not exist real polynomials $f(x)$ and $g(x)$ with higher degrees such that $(x^2+A)f(x)^2+1\equiv g(x)^2$. Any help will be appreciated.

Answer 1

This is not really a question about sequences but about polynomials (you are asking when a polynomial has a square root and this can be understood in terms of its irreducible factorization), and, you also do not need to answer the question you've asked in order to solve the problem you're working on. Your desired identity can be written as a Pell equation

$$g(x)^2 - (x^2 + A) f(x)^2 = 1$$

which means the machinery of Pell equations can be used to find infinitely many solutions, adapted to this case. The idea is that we are going to factor the LHS as

$$(g(x) + f(x) \sqrt{x^2 + A})(g(x) - f(x) \sqrt{x^2 + A}) = 1.$$

More precisely, we are going to consider expressions of the form $\alpha = g(x) + f(x) \sqrt{x^2 + A}$, which can be added and multiplied to produce other expressions of the same form (this is an extension $\mathbb{R}[x, \sqrt{x^2 + A}]$ of $\mathbb{R}[x]$). The conjugate of such a thing is $\overline{\alpha} = g(x) - f(x) \sqrt{x^2 + A}$, and the norm is

$$N(\alpha) = \alpha \overline{\alpha} = g(x)^2 - (x^2 + A) f(x)^2.$$

Then the problem is equivalent to finding elements of norm $1$. The reason to restate the problem this way is that the norm has a crucial property:

Proposition: The norm is multiplicative: $N(\alpha \beta) = N(\alpha) N(\beta)$.

This is a nice exercise, concretely it is a version of Brahmagupta's identity and abstractly it follows from the observation that the conjugate is multiplicative: $\overline{\alpha \beta} = \overline{\alpha} \overline{\beta}$. This means we can find elements of norm $1$ by taking powers of other elements of norm $1$, or by multiplying two elements of norm $-1$. For example, taking $g(x) = x, f(x) = 1$ gives

$$N(x + \sqrt{x^2 + A}) = -A$$

so $\alpha = \frac{x}{\sqrt{A}} + \frac{\sqrt{x^2 + A}}{\sqrt{A}}$ is an element of norm $-1$. This means that the sequence

$$\alpha^{2k} = \left( \frac{x}{\sqrt{A}} + \frac{\sqrt{x^2 + A}}{\sqrt{A}} \right)^{2k}$$

is an infinite sequence of elements of norm $1$, so its components give infinitely many solutions to the Pell equation of arbitrarily high degree. They can be computed inductively by taking powers $\alpha^2 = \frac{2x^2 + A}{A} + \frac{2x}{A} \sqrt{x^2 + A}$ (which is your solution). For example, the next term of the sequence after $\alpha^2$ is

$$\alpha^4 = \frac{8x^4 + 8x^2 A + A^2}{A^2} + \frac{8x^3 + 4Ax}{A^2} \sqrt{x^2 + A}$$

which gives the solution $g(x) = \frac{8x^4 + 8x^2 A + A^2}{A^2}, h(x) = \frac{8x^3 + 4Ax}{A^2}$. Generally there is also machinery in the theory of Pell equations to prove you've found all solutions but it's usually used over $\mathbb{Z}$ and I don't know off the top of my head how to adapt it to this case.

Edit: Okay, I think we can argue that the above procedure produces all solutions up to sign (which means $\alpha^2$ is what is called a fundamental unit). The idea is to argue that given a solution we can produce a solution of smaller degree by multiplying by $\alpha^2$ or $\alpha^{-2}$.

Answer 2

You want two polynomials $f(x)$ and $g(x)$ with real coefficients so that

$$ (x^2 + A) f(x)^2 + 1 = g(x)^2$$ i.e. $$ f(x)^2 = \frac{g(x)^2-1}{x^2 + A} = \frac{(g(x)+1)(g(x)-1)}{x^2 + A} $$ The roots of $x^2 + A$ are $\pm i \sqrt{A}$, so these must be roots of $g(x)+1$ or $g(x)-1$. That is, $g(i\sqrt{A}) = \pm 1$. Since the coefficients of $g$ are real, $g(-i\sqrt{A}) = \overline{g(i \sqrt{A})} = g(i \sqrt{A})$. If $g(i\sqrt{A}) = g(-i\sqrt{A}) = 1$, $g(x)-1$ is divisible by $x^2+A$, i.e. for some polynomial $h$ with real coefficients, $g(x) = 1 + (x^2 + A) h(x)$, and your equation becomes $$ f(x)^2 = ((x^2 + A) h(x) + 2) h(x) $$ Now the two factors on the right are coprime, so both are squares of polynomials (though not necessarily with real coefficients). Let's write $h(x) = p(x)^2$ and $(x^2 + A) p(x)^2+2 = q(x)^2$. Note that if $p$ has degree $k$, $q$ has degree $k+1$.

Some solutions: $$\eqalign{ p(x) &= \frac{2}{\sqrt{-2A}},\ q(x) = \frac{2x}{\sqrt{-2A}},\ h(x) = -\frac{2}{A},\cr f(x) &= \frac{2}{A} x,\ g(x) = 1 - \frac{2(x^2+A)}{A}\cr p(x) &= \frac{2 \sqrt{2} x}{A},\ q(x) = -\frac{2 \sqrt{2}x^2}{A} - \sqrt{2},\ h(x) = \frac{8 x^2}{A^2},\cr f(x) &= \frac{4 x (2 x^2 + A)}{A^2}, \ g(x) = 1 + \frac{8 (x^2 + A) x^2}{A^2}\cr p(x) &= -\frac{8x^2}{A \sqrt{-2A}} - \frac{2}{\sqrt{-2A}}, q(x) = \frac{8 x^3}{A \sqrt{-2A}} + \frac{6x}{\sqrt{-2A}},\cr h(x) &= - \left(\frac{8 x^2}{\sqrt{2 A^3}} - \frac{2}{\sqrt{2A}} \right)^2, \cr f(x) &= \frac{2 (4 x^2 + A) (4 x^2 + 3 A) x}{A^3},\cr g(x) &= 1 - (x^2 + A) \left( \frac{8 x^2}{2 A^3} + \frac{2}{\sqrt{2 A}}\right)^2} $$