details, details. From $x$ odd and

$$ x^4-32x-16 = (x^2 + 4)^2 - 2(2x+4)^2. $$ we see that $x^4-32x-16$ is not divisible by any prime $q \equiv 3,5 \pmod 8.$ That is, $x^2 + 4$ is also odd. Next, $\gcd(x^2 + 4, 2x+4) =\gcd(x^2 + 4, x+2).$ However, $(x+2)(x-2) = x^2 - 4,$ so that $\gcd(x^2 + 4, x+2)$ divides $8$ and must be $1.$ Thus the representation $x^4-32x-16 = (x^2 + 4)^2 - 2(2x+4)^2$ is primitive.

Now, $x^4 - 32 x - 16$ is often divisible by primes $7 \pmod 8$ and are therefore ruled out. Next, it appears that $x^4 - 32 x - 16$ has always an odd number of prime factors of shape $4 u^2 + 4uv + 9 v^2,$ and I am fiddling with showing that these numbers are a dead end as well. This could require understanding every word in Liu and Williams, hard to say

details, details. From $x$ odd and

$$ x^4-32x-16 = (x^2 + 4)^2 - 2(2x+4)^2. $$ we see that $x^4-32x-16$ is not divisible by any prime $q \equiv 3,5 \pmod 8.$ That is, $x^2 + 4$ is also odd. Next, $\gcd(x^2 + 4, 2x+4) =\gcd(x^2 + 4, x+2).$ However, $(x+2)(x-2) = x^2 - 4,$ so that $\gcd(x^2 + 4, x+2)$ divides $8$ and must be $1.$ Thus the representation $x^4-32x-16 = (x^2 + 4)^2 - 2(2x+4)^2$ is primitive.

Now, $x^4 - 32 x - 16$ is often divisible by primes $7 \pmod 8$ (with odd exponent) and are therefore ruled out. Next, it appears that $x^4 - 32 x - 16$ has always an odd number of prime factors of shape $4 u^2 + 4uv + 9 v^2,$ and I am fiddling with showing that these numbers are a dead end as well. This could require understanding every word in Liu and Williams, hard to say