# solutions of gold APN functions using trace function

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?

• Please modify your question specifying where and how the trace comes in, it's incomplete as it stands. – kodlu Jan 15 at 0:11
• $g(x) = x^{2^k}$ 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 $x$ – reuns Jan 15 at 3:21