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Given that $n\in\mathbb{N}$. How many pairs of $P(x),\;Q(x)$ (all coefficients of $P$ and $Q$ are real numbers) with $deg\; P > deg\; Q$ satisfying: $[P(x)]^2+[Q(x)]^2=x^{2n}+1$?

From the condition $deg\; P > deg\; Q$ we can say $\text{deg}(P)=n$. Also, $x^{2n}+1$ has $2n$ roots (complex and real) hence $[P(x)]^2+[Q(x)]^2$ too has $2n$ roots. Let $z, \bar{z}$ be a pair of complex roots of $x^{2n}+1$, then $$\begin{array}{l} (P(z))^2+(Q(z))^2=0 \\ (P(\bar{z}))^2+(Q(\bar{z}))^2=0 \end{array}$$ Also, we can rewrite the given expression as $$[P(x)+iQ(x)][P(x)-iQ(x)]=(x^n-i)(x^n+i)$$ So, $$P(z)=\pm iQ(z), P(\bar{z})=\pm iQ(\bar{z})$$ Now $$\deg(P(z))=n=\deg(\pm iQ(z))$$ Similarly $$\deg(P(\bar{z}))=n=\deg(\pm iQ(\bar{z}))$$ Where do I proceed from here? Please only provide hints and not solutions until I ask (if after I have tried everything and still not able to solve the problem). Thanks.

Yoda_2008
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  • I don't think $P(z) = \pm i Q(z)$ is a correct conclusion here? It would mean that $x^{2n}+1=0$. – Oleksandr Kulkov Apr 09 '23 at 18:08
  • As a hint, consider all roots of $x^{2n}+1$ and think on how you can group them together to make $P(x)$ and $Q(x)$. – Oleksandr Kulkov Apr 09 '23 at 18:10
  • @OleksandrKulkov yes I'm considering $z, \bar{z}$ to be the roots of $x^{2n}+1$. So, it would make it 0, that's the intention. – Yoda_2008 Apr 09 '23 at 18:10
  • Oh, sorry, I misread this part :)

    Then, consider the full factorization of $P^2(x) + Q^2(x)$ into linear monomials.

     – Oleksandr Kulkov Apr 09 '23 at 18:12
  • @OleksandrKulkov ok working with all the roots of $x^{2n}+1$ is tricky. Because they would include both reals and complex roots. Also can you elaborate about grouping them to maje $P(x)$ and $Q(x)$. – Yoda_2008 Apr 09 '23 at 18:15
  • @Yoda_2008 The problem asks for $,\deg P \gt \deg Q,$, so you can't claim $,\deg P = \deg Q,$. For an example, consider $,(x^3-2x)^2+(2x^2-1)^2$ $= x^6+1,$. – dxiv Apr 09 '23 at 18:50
  • @dxiv then where did my argument went wrong? Like I assumed $z$ to be one of the roots of $x^{2n}+1$, hence $$[P(z)+iQ(z)][P(z)-iQ(z)]=0$$ right? Which follows $$P(z)=\pm iQ(z)$$ which step is wrong – Yoda_2008 Apr 09 '23 at 18:56
  • @Yoda_2008 The first equality only holds for the $2n^{th}$ roots of $-1$, and the second one doesn't follow from that as a polynomial equality. I suggest you try it on the example from my previous comment and see why/where it breaks down. – dxiv Apr 09 '23 at 18:59
  • Similar question (with integer coefficients) https://artofproblemsolving.com/community/c6h1807765p12028810 – Sil Apr 09 '23 at 19:06
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    $P(z) = \pm i Q(z)$ doesn't imply $\deg P = \deg Q$, as long as $z$ is concrete number rather than a variable. – Oleksandr Kulkov Apr 09 '23 at 19:58

2 Answers2

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Another strategy: Factor $x^{2n}+1$ as product of $n$ quadratic polynomials with roots $z_k=a_k\pm b_ki$, represent each as sum of squares $(x-a_k)^2+{b_k}^2$, and multiply sum of square representations together with the Brahmagupta-Fibonacci identity. This should give $2^n$ representations.

nor
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Inserting integers $k=0,\ldots,n-1$ in $P(x)^2+Q(x)^2=x^{2n}+1$ results in $n$ equations $$P(k)^2+Q(k)^2=k^{2n}+1.$$ Each of these equations is a representation of $k^{2n}+1$ as a sum of squares. The number of representations $c=a^2+b^2$ of a given $c$ depends on the prime divisors of $c$, as known from Number Theory. Each sequence of representations $$k^{2n}+1=a_k^2 + b_k^2$$ results by interpolation $P(k)=a_k$ in a unique polynomial $P(x)$.

Gary
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nor
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  • How do you know $P(k),Q(k)$ are integers? OP stated $P(x),Q(x)$ are polynomials with real coefficients. – Sil Apr 10 '23 at 13:49
  • @Sil you are right. I have overlooked this in the OP, sorry for my mistake. – nor Apr 10 '23 at 14:00