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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?

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    $\begingroup$ Please modify your question specifying where and how the trace comes in, it's incomplete as it stands. $\endgroup$ – kodlu Jan 15 at 0:11
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    $\begingroup$ $g(x) = x^{2^k}$ is a $GF(2)$-linear map so $F(x+a) = (x+a)g(x+a) = (x+a)(g(x)+g(a))$ and $F(x)+F(x+a) = ag(x)+ag(a) $ which is an affine map in $x$ $\endgroup$ – reuns Jan 15 at 3:21

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