1

Fix any field, for concreteness the complex numbers. Let $a$ be any element of the field. Are there any non-constant rational functions $f$ over the field such that $f(x)^4+a$ is a perfect square? That is, $f(x)^4+a = r(x)^2$, where $r(x)$ is another rational function?

Equivalently, do there exist any pairs of polynomials $p,q$ such that $p(x)^4+aq(x)^4$ is a perfect square?

For the related problem of finding $f(x)$ such that $f(x)^2+a$ is a perfect square, a solution is $f(x)=(4x^2-a)/(4x)$. Notice that $$f(x)^2+a=\frac{(4 x^2-a)^2+16x^2a}{16 x^2}=\frac{(4 x^2+a)^2}{16 x^2}$$ which is a perfect square.

Is there a way to generalize this to an $f$ such that $f(x)^4+a$ is a perfect square? Alternatively, is there a way to argue that such a rational function does not exist?

vitamin d
  • 5,913
AAA
  • 857

2 Answers2

1

Complex numbers
In the case of $a=-1$(or any negetive forth power), in your polynomial form, it implies there exist $a,b,c$ non-zero such that $$a^4-b^4=c^2$$ contradicting Fermat’s right triangle theorem
And in the case of $a=1$, or any positive forth power, it implies the existence of solutions to the diophantine equation $$a^4+b^4=c^2$$ which doesn’t have non-trivial solutions.
See: Pythagorean triples and perfect squares
finite fields
if $a\ne0$ then such a function exist if there exist two elements $b,c$ that are quatric residues and such that $b-a,c-a$ are quartic residues too.
$f(x)=\begin{cases} \root 4\of {b-a} & x=0\\ \root 4\of {c-a}& x\ne0 \end{cases}$
because Every Function in a Finite Field is a Polynomial Function this is a solution.

In $GF(2^n)$
remember that $GL(2^n)$ is of charcteristic 2, and thus $x=-x$.
if $a^2=b^2$ then $a^2-b^2=0$ which can be factorized to $(a-b)(a+b)=(a+b)^2=0$ and because it is a field, $a=b$, so all squares are unique and thus all numbers have square roots.(remember the field is finite)
and thus all numbers have square roots so do they have quartic roots.
and thus $f(x)=\begin{cases} \root 4\of {1-a} & x=0\\ \root 4\of {a}& x\ne0 \end{cases}$ solves it.
In $GF(p)$ ($p\ne 2$)
from Quartic reciprocity:
at least one of $r,-r$ is a quartic residue.
if $-a$ is a quartic residue then we have $0,0-a$ both quartic residues.
if $a$ is a quartic residue then we have $a,a-a$ both quartic residues.
It isn't always possible though, for example in $GF(3)$ and $a=2$ only $1,0$ are quartic residues, and a function is impossible.

razivo
  • 2,285
  • This would only be a contradiction for integer $a,b,c$. I suppose by multiplying out the denominators, it would also be a contradiction for rational constants $a,b,c$. But this wouldn't seem to give a contradiction for algebraically closed fields. – AAA Jul 12 '21 at 19:40
  • It just means that if it does exist for the complex numbers, which I doubt it does, it probably won't look like the square one, as it can't work for negative forth power $a$'s – razivo Jul 12 '21 at 20:11
  • For GF(p), are you assuming $r$ is already a quadratic residue? Otherwise the claim that one of $r,-r$ is a quartic residue is false if $p\equiv 1\pmod 4$. – AAA Jul 23 '21 at 05:29
1

There do not exist such $f$, for any non-zero $a$, provided the field characteristic is not 2.

It turns out this follows from the answer to another question of mine (Characterize polynomials that can be "squarified").

Over a closed field (of characteristic not 2), $-a$ will have 4 distinct roots. Hence, the polynomial $p(y)=y^4+a=y^4-(-a)$ is square-free. The answer to my other question shows that any square-free polynomial of degree greater than 2 cannot be "squarified". I.e. there do not exist rational functions $f(x)$ and $r(x)$ such that $r(x)^2=p(f(x))=f(x)^4+a$.

AAA
  • 857