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Suppose you have two polynomials $p,q$ with the guarantee that their composition $p(q(x))$ is a perfect square, i.e. that there exists a polynomial $r$ with $p(q(x))=r(x)^2$.

Can we characterize such pairs of polynomials?

Conjecture: Over any algebraically closed field, all such $p,q$ have one of two forms:

  1. $p$ is a perfect square.
  2. $q(x)-c$ is a perfect square for some constant $c$, and $p(y)=(y+c)\times s(y)^2$ for some polynomial $s$.

Clearly any $p,q$ satisfying 1 or 2 will have $p(q(x))$ be a perfect square. Are there any other examples, or can we prove that 1 or 2 are exhaustive?

AAA
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  • Related SOS polynomials – Surb Jun 01 '21 at 22:16
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    You also have the case that $q$ is constant and $p$ arbitrary. – plop Jun 01 '21 at 22:18
  • Adding the case I mentioned. You can assume that $p$ is square-free, since the square part already forms a square. Let $p(x)=(x-a_1)...(x-a_n)$, with $a_i$ different roots. If $n<2$ we are done. Assume $n\geq2$. Observe that $q(x)-a_i$ are relatively prime. Therefore, for $p(q(x))$ to be a square, each $q(x)-a_i$ must be a square. But then $q(x)-a_1=r^2(x)$ and $q'(x)=2r(x)r'(x)$ have a common factor of positive degree equal to half the degree of $q$. But then $q(x)-a_2=q(x)-a_1+(a_1-a_2)$ must also have the a common factor with $q'(x)$ of the same degree... – plop Jun 01 '21 at 22:21
  • ... But this implies that $q(x)-a_1$ and $q(x)-a_2$ have a common factor of positive degree, which is a contradiction. Therefore, $p(x)$, or rather the square-free part of the original $p(x)$, is of degree $<2$. – plop Jun 01 '21 at 22:22
  • @CalvinLin I didn't say they are the same factor. $q'$ has degree $2n-1$ and $q(x)-a_1$ shares with $q'$ a factor of degree $n$. That is more than half the degree of $q'$. Likewise $q(x)-a_2$ shares with $q'$ a factors of degree $n$. Therefore, $q(x)-a_1$ and $q(x)-a_2$ share a factor of positive degree. – plop Jun 01 '21 at 23:14
  • @CalvinLin The subject of that sentence is "$q(x)-a_1=r^2$ and $q'(x)=2r(x)r'(x)$". They have as common factor $r$ and $r$ has degree equal to half of the degree of $q$. Then the next sentence repeats the claim for $q(x)-a_2$. Finally the next comment, concludes from that that $q(x)-a_1$ and $q(x)-a_2$ must have a common factor of positive degree. – plop Jun 01 '21 at 23:21
  • @CalvinLin The argument is not using that $q(x)-a_1=q(x)-a_2$. The common factor that each of them have is with $q'$, which is the same derivative for both. Each of them has a factor in common with $q'$ that is of degree $\deg(q)/2$. Therefore, those common factors have in turn a factor in common. – plop Jun 01 '21 at 23:49
  • @CalvinLin With formulas: Let $\deg(q)=2n$. Then $\deg(\gcd(q-a_i,q'))=n$, for $i=1,2$, since $q-a_i$ are squares. Now, $\deg(q')=2n-1<n+n=\deg(\gcd(q-a_i,q'))+\deg(\gcd(q-a_i,q'))$. Therefore, $\deg(\gcd(q-a_1,q-a_2))>0$. – plop Jun 01 '21 at 23:57
  • @plop I'm just going to consider this a miscommunication / that I misinterpreted what you're saying. – Calvin Lin Jun 02 '21 at 00:13
  • @plop Excellent, thanks! – AAA Jun 03 '21 at 21:55

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