Let $a+bc=x^2, b+ac=y^2, c+ab=z^2$. It suffices to show that $\nu_p((a+b)(b+c)(c+a))$ is even for any prime $p \equiv 3 \pmod 4$. We will show that this is true for all terms $a+b, b+c, c+a$; focus on $a+b$ and suppose that $\nu_p(a+b)$ is odd for some prime $p \equiv 3 \pmod 4$; in particular, $p \mid a+b$. We have that $a+bc \equiv a(1-c) \pmod p$ and $b+ac \equiv b(1-c) \pmod p$ are quadratic residues modulo $p$, so, in particular, $ab(1-c)^2$ is a quadratic residue modulo $p$. We have the following three cases: $\textbf{Case 1.}$ None of $a, b, 1-c$ are divisible by $p$. Then $ab \equiv -a^2$ is a quadratic residue modulo $p$, i.e. $-1$ is a quadratic residue, contradiction. $\textbf{Case 2.}$ We have $c \equiv 1 \pmod p$. Then $x^2+y^2=a+bc+b+ac=(a+b)(c+1)$ and $p \nmid c+1$, so $\nu_p(x^2+y^2)=\nu_p(a+b)$ is even, contradiction. $\textbf{Case 3.}$ We have $p \mid a$ and $p \mid b$. If $c \neq - 1 \pmod p$, then similarly as above $\nu_p(x^2+y^2)=\nu_p(a+b)$ is even. Otherwise, $c+ab \equiv -1 \pmod p$ is not a quadratic residue, contradiction.

Note that $$a^2(b+c) + b^2(c+a) + c^2(a+b) + 2abc = (a+b)(b+c)(c+a),$$so it suffices to show that $(a+b)(b+c)(c+a)$ can be expressed as the sum of two squares. In fact, we will show the stronger statement that $a+b, b+c$, and $c+a$ can each be expressed as the sum of two squares. To prove this stronger statement, we will WLOG show that $a+b$ can be expressed as the sum of two squares, since the other expressions being the sum of two squares follows by symmetry. First assume that $a+bc=x^2$, $b+ca=y^2$, and $c+ab=z^2$ for positive integers $x,y$, and $z$. Then, note that from adding the first two of these three equations, we have that $$(a+b)(c+1) = x^2+y^2.$$Now, it is well known that $n$ being the sum of two squares is equivalent $2\mid \nu_p(n)$ for every $p\equiv 3\pmod{4}$. Now, consider any $p\equiv 3\pmod{4}$ prime, I claim that $p\nmid \gcd(a+b, c+1)$. Assume the contrary, then $c\equiv -1\pmod{p}$ and $a\equiv -b\pmod{p}$. Note that this means $$x^2 \equiv a-b \pmod{p},$$$$y^2 \equiv b-a\pmod{p},$$$$z^2\equiv ab-1\pmod{p}.$$Therefore, $p\mid x^2+y^2$, but since $p\equiv 3\pmod{4}$, $-1$ is not a QR mod $p$, so it follows that $x\equiv y\equiv 0\pmod{p}$. Then, $a-b\equiv 0\pmod{p}$, so $a\equiv b\pmod{p}$. Now, since $p\mid a+b$, it follows that $p\mid 2a$, i.e. $a\equiv 0\pmod{p}$. Then, $b\equiv 0\pmod{p}$, but then $$z^2 \equiv ab-1 \equiv -1\pmod{p},$$contradicting the fact that $-1$ is an NQR mod $p$ (as $p\equiv 3\pmod{4}$). Therefore, we must have that $p\nmid \gcd(a+b, c+1)$, as claimed. Now, this means that for every $p\equiv 3\pmod{4}$ prime dividing $a+b$, we have that $$\nu_p(x^2+y^2) = \nu_p(a+b) + \nu_p(c+1) = \nu_p(a+b),$$since if $\nu_p(a+b) > 0$, then we cannot have $\nu_p(c+1) > 0$ as well else $p$ would divide the gcd (which we just showed could not happen). Since $x^2+y^2$ is the sum of two squares, $\nu_p(x^2+y^2)$ is even, so $\nu_p(a+b)$ is even. This holds for all $p\equiv 3\pmod{4}$ dividing $a+b$, so $a+b$ can be expressed as the sum of two squares. As discussed earlier, similar arguments imply that $b+c$ and $c+a$ can be expressed as the sum of two squares, so it follows that $(a+b)(b+c)(c+a)$ can be expressed the sum of two squares (it's well known that if $r$ and $s$ are expressible as the sum of two squares, then so is $rs$), as desired. $\blacksquare$

The equation simplifies to $(a+b)(b+c)(c+a)$. The problem is now equivalent to showing $\nu_p(a+b)$ is even. Let $a+bc=x^2$ and define $y^2,z^2$ similarly. Now FTSOC take $p=4k+3$ such that $\nu_p(a+b)$ is odd. Then $p\mid a+b$ and $\nu_p(x^2+y^2)=\nu_p((a+b)(c+1))$ should be even or $p\mid c+1$. So $x^2\equiv a-b$ and $y^2\equiv b-a$ mod $p$ and they are also QRs. Thus, $a\equiv b \equiv 0 \pmod{p}$ since $p\mid a+b$. But this gives us $x^2\equiv -1\pmod{p}$ which is not a QR mod $p$. Hence $\nu_p(a+b)$ is even and we are done! $\blacksquare$