Can someone check this solution plz, thanks. Here we suppose that polynomial $A\in \mathbb{R}[x]$ divisible by polynomial $B\in \mathbb{R}[x]$ iff there exist polynomial $C\in \mathbb{R}[x]$ that $A(x)=B(x)C(x)$ We will prove the statement by induction on $n\in \mathbb{Z}^+$ For $n=1$, let $P(x)=(x^2+1)P_1(x)+ax+b$ where $a,b\in \mathbb{R},P_1\in \mathbb{R}[x]$ (i.e. Divide $P(x)$ by $x^2+1$) We need to show that there exist $Q\in \mathbb{R}[x]$ that $(ax+b)^2+Q(x)^2$ divisible by $x^2+1$ Choose $Q(x)=-bx+a$, we get that $(ax+b)^2+Q(x)^2=(x^2+1)(2a^2+2b^2)$ For induction step, suppose the statement is true for $n\in \mathbb{Z}^+$ We will prove that it's true for $n+1$ For any polynomial $P\in \mathbb{R}[x]$, by induction hypothesis, there exist polynomial $T,S\in \mathbb{R}[x]$ that $P(x)^2+T(x)^2=(x^2+1)^nS(x)$ If $T(x)$ is divisible by $x^2+1$, we get that $P(x)$ is divisible by $x^2+1$ Then we need to show that there exist $Q\in \mathbb{R}[x]$ that $\Big( \frac{P(x)}{x^2+1}\Big)^2+Q(x)^2=(x^2+1)^{n-1}H(x)$ for some $H\in \mathbb{R}[x]$ which is the induction hypothesis when $n\Leftarrow n-1$ So suppose that $T(x)$ is not divisible by $x^2+1$ Let $Q(x)=T(x)+(x^2+1)^nU(x)$ for some polynomial $U\in \mathbb{R}[x]$ We get $P(x)^2+Q(x)^2=(x^2+1)^nS(x)+2T(x)U(x)(x^2+1)^n+(x^2+1)^{2n}U(x)^2$ Thus, we need to show that there exist $U\in \mathbb{R}[x]$ that $S(x)+2T(x)U(x)$ is divisible by $x^2+1$ Let $T(x)=(x^2+1)M(x)+wx+z$ where $w,z\in \mathbb{R},M\in \mathbb{R}[x]$ and $w,z$ are not both zero. And let $U(x)=S(x)V(x)$, we need to show that there exist polynomial $V\in \mathbb{R}[x]$ that $1+2(wx+z)V(x)$ divisible by $x^2+1$ Choose $V(x)\equiv \frac{w}{2z^2+2w^2}x-\frac{z}{2z^2+2w^2}$, it's not hard to verify that this polynomial satisfy the condition. This complete induction step and we are done.

There's also a slick algebraic solution using complex numbers (from the Kömal website). Clearly, $P$ is only there for aesthetic purposes - we just need a $Q$ for which $1+Q^2$ is divisible by $(1+x^2)^n$. Let $M(x)=(x^2+1)^n=M^+(x)M^-(x)$, where \begin{align*} M^+(x)&=(x+i)^n=A(x)+B(x)i,\\ M^-(x)&=(x-i)^n=A(x)+B(x)i \end{align*}for real polynomials $A,B$. We want some $Q$ with \[ M=A^2+B^2|1+Q^2, \]for which it suffices to guarantee $A+Bi|1+Qi$ (as then the complex conjugate holds too), for which it suffices to choose some $C,D$ with \[ (A+Bi)(C+Di)=1+Qi, \]for which it suffices to make $AC-BD=1$, which is possible due to Bézout's lemma, as $A,B$ are coprime (they clearly cannot share a root, as then $M^+$ and $M^-$ would share a root). Comment: this solution generalizes to any $M$ with no real root effortlessly (well, the inductive one does, too, using the Chinese Remainder Theorem for polynomials). How would it generalize to other polynomials rings, though?

Edit: $K$ must also be a perfect field (nearly all relevant fields are, and this does not affect the solution to the original problem). randomusername wrote: Comment: this solution generalizes to any $M$ with no real root effortlessly. How would it generalize to other polynomials rings, though? In fact, $\forall n,\ \exists Q_n\in \mathbb R[x]$ satisfying $M^n\mid 1+Q_n^2$ if and only if $M$ has no real roots. Actually, we can prove a much more general Claim: Let $K$ and $\text{ prime }M\in K[x]$ be a field and irreducible polynomial of our choosing. Furthermore, let $\text{ prime }P\in K[x]$ and $P\in \left\{K[x]\right\}[R]$ be two polynomials whose corresponding coefficients are the "same" in that the former's are just elements of $K$, and the latter's are said elements of $K$ acting as polynomials in $K[x]$. (They are so similar, I won't bother giving them different names). Then $$\exists Q\in K[x]\text{ s.t. }P(Q)\underbrace{=}_{K[x]/(M)}0\iff\exists \omega \in \overline{K},\ \nu\in K(\omega)\text{ s.t. }M(\omega)\underbrace{=}_{K}0\text{ and }P(\nu)\underbrace{=}_{K(\omega)}0$$Proof: A well-known field theorem dictates that $K[x]/(M)$ is a field, and in particular that it is isomorphic to $K(\omega)$. In fact, we can be explicit in constructing (at least the forward half of) said isomorphism; take the map $\eta:K[x]/(M)\to K(\omega)$ given explicitly by $\eta:R\mapsto R(\omega)$. In particular by corresponding through $\eta$ and $\eta^\text{op}$, we see that $P\in K[x]\subseteq K(\omega)[x]$ and $P\pmod{M}\in \left\{K[x]/(M)\right\}[R]$ map to each other, and thus a root of the former polynomial over $K(\omega)$ when taken through $\eta^\text{op}$ yields a root of the latter over $K[x]/(M)$ and visa versa $\Box$ Now we will show that the original problem is in fact implied by Claim. Of course, we start by recreating Hensel's Lemma: If $P\in \left\{K[x]\right\}$ has a root $R_n\in K[x]/\left(M^n\right)$, then there is a unique $R_{n+1}\in K[x]/\left(M^{n+1}\right)$ which is also a root of $P$, and that furthermore satisfies $M^n\mid R_{n+1}-R_n$. Proof: Exactly the same as that of Hensel's Lemma. In particular, the perfection of $K$ implies that $P\in K[x]\subseteq K(\omega)[x]$ is squarefree (but not necessarily prime in the latter ring), and $\eta^\text{op}$ ports this squarefreeness over to $\left\{K[x]/(M)\right\}[R]\ \Box$ From now the path is clear: A direct analogue for CRT exists over the quotient rings of $K[x]$, and thus taken in combination with the above, we see that Claim is satisfied for all $M\in K[x]$ iff it is satisfied for all the of the prime factors of $M$ (the "only if" arises by simply taking quotients by the ideals generated by said prime factors). Henceforth take $K=\mathbb R\text{ and }$ $P(\ast):=1+\ast^2$. In this way the "no real roots" condition becomes clear, since a real root would imply a degree-one prime factor $x-\omega$, which would in turn contradict Claim, since $\mathbb R(\omega)=\mathbb R$. Alternatively, a lack of real roots and the fact that $[\mathbb C:\mathbb R]=2$ would mean that any root $\omega$ of a prime factor of such an $M$ would yield $\mathbb R(\omega)=\mathbb C$, in accordance with Claim $\blacksquare$

And the solution of that Kömal problem also can apply to any $M(x)$ having no real roots. In this case $M(x)=(1+x^2)^n$