The Gold APN is defined as $F(x)=x^{2^{k}+1}$ in $GF(2^n)$, where $\gcd(k,n)=1$. The differential uniformity computed using $F(x)=F(x+a)=b$ as following:
$x^{2^{k}+1} + (x+a)^{2^{k}+1}=b$
$x^{2^{k}+1} + (x+a)^{2^{k}}(x+a)=b$
$x^{2^{k}+1} + (x^{2^k}+a^{2^k})(x+a)=b$
$x^{2^{k}+1} + x^{2^{k}+1} +x^{2^k}a +a^{2^k}x +a^{2^{k}+1} =b$
$x^{2^k}a +a^{2^k}x =b +a^{2^{k}+1}$
dividing both sides by $a^{2^k+1}$
$x^{2^k}(a^{-1})^{2^k}+xa^{-1}=b(a^{2^k+1})^{-1}+1$
from this point onward i got stuck to prove that the Gold APN has two solutions using trace functions.
if solution exists
$tr(x^{2^k}(a^{-1})^{2^k}+xa^{-1})=0=tr(b(a^{2^k+1})^{-1}+1)$
Q1: How to apply the trace function to find the roots of the differential uniformity function?