Prove that $(d (a + e) + b (c + e) + (a + c) f)^2 - 4 (a c + a e + c e) (b d + b f + d f)$ cannot be factored as a square

Question

For understanding of the question let's start with two examples:

We can factor $V(a,b,c)=a^4 + 2 a^2 b + b^2 - 2 a^2 c - 2 b c + c^2$ as a square, i.e we can write $V$ as square of a function $p(a,b,c)=a^2 + b - c$ so that holds $V=p^2$. Another example is $W(a,b)=a^2 + 2 a b + 2 a^{3/2} b + b^2 + 2b^2 \sqrt a + a b^2$ that can be factored as $W=r^2$ with $r(a,b)=a + b\sqrt a + b$.

The problem:

The function of interest is $U(a,b,c,d,e,f)=(d (a + e) + b (c + e) + (a + c) f)^2 - 4 (a c + a e + c e) (b d + b f + d f)$

with $(a,b,c,d,e,f)\in \mathbb{R}$. The aim is to take the square root of $U$. Therefore I tried to factor $U$ as a square, but failed, i.e. could not find the function $q(a,b,c,d,e,f)\in \mathbb C$ with $U=q^2$. How to prove that factorization is (not) possible? For factorization I consider for $q$ any mathematical function.

Answer 1

We can factor $V(a,b,c)=a^4 + 2 a^2 b + b^2 - 2 a^2 c - 2 b c + c^2$ as a square

Considering it as a polynomial $P(a)$ in $a$ with coefficients parameterized in $b,c$, the condition for a multiple root is for $\gcd(P(a), P'(a))$ to have degree $\ge 1$. Calculating the polynomial $\gcd$ gives (verified by WA):

$$ \gcd(a^4 + 2 a^2 b + b^2 - 2 a^2 c - 2 b c + c^2, 4 a^3 + 4 a b - 4 a c) = a^2 + b - c $$

Therefore $(a^2 + b - c)^2 \,|\, P(a)$, and in fact $(a^2 + b - c)^2 \,=\, P(a)\,$ by degree considerations.

$U(a,b,c,d,e,f)=(d (a + e) + b (c + e) + (a + c) f)^2 - 4 (a c + a e + c e) (b d + b f + d f)$

Considering it as a polynomial $Q(a)$ in $a$, it can be verified that $\gcd(Q(a), Q'(a)) = 1$ (WA) so $Q(a)$ has no polynomial factor at a power greater than $1$, which implies that it is not the square of another polynomial.

Answer 2

We have $$ U(1,-2,1,3,1,0)=76, $$ which is not a square in $\Bbb Z$. Suppose that $U\in \Bbb Z[a,b,c,d,e,f]$ is a square, i.e., $$U(a,b,c,d,e,f)=(p(a,b,c,d,e,f)^2$$ for some polynomial $p\in \Bbb Z[a,b,c,d,e,f]$ - this is meant by a factorization. Then every integer value of $U$ needs to be a square in $\Bbb Z$. This is not true, as $76$ is not a square.

On the other hand, $V=a^4 + 2 a^2 b + b^2 - 2 a^2 c - 2 b c + c^2$ is a square of a polynomial $p\in \Bbb Z[a,b,c]$, namely $p(a,b,c)=a^2+b-c$. And all integer values for $V$ are indeed squares.

Answer 3

If you consider $U$ as a function $\mathbb R^6 \to \mathbb R$, then it is a square of another function $q: \mathbb R^6 \to \mathbb C$. Note that we need complex numbers in the codomain here, but cannot also take them in the domain! Define $q$ by $$q(a,b,c,d,e,f) = \begin{cases} \sqrt{U(a,b,c,d,e,f)} & \text{if} & U(a,b,c,d,e,f) > 0 \\ i \sqrt{-U(a,b,c,d,e,f)} & \text{if} & U(a,b,c,d,e,f) < 0\end{cases}$$ where $i \in \mathbb C$ is the imaginary unit. Then clearly $$q(a,b,c,d,e,f)^2 = U(a,b,c,d,e,f)$$ for all $a,b,c,d,e,f \in \mathbb R$.