$$ \text{Let}\space A:= \left\{\space x^p+y^p=z^p\in\mathbb{N}^* \space\space\middle|\space\space \begin{array}{ll} x,y,z,p\in\mathbb{N}^*\space\text{such that p is prime and} \\ \forall x_0,y_0,z_0\in\mathbb{N}^*\space\text{such that}\space x_0^p+y_0^p=z_0^p\in\mathbb{N}^*, \\ \text{if}\space x_0y_0z_0\ne xyz\space\text{then}\space xyz<x_0y_0z_0 \end{array} \space \right \}. \\ $$
$A\subset\mathbb{N}^*.$
$$\forall a,b,c\in (A\cup\{0\})\space\text{and}\space d,g,h\in(\mathbb{N}^*\cup\{0\})\space\text{let}\space(a,d),(b,g),(c,h)\in (A\cup\{0\})\times(\mathbb{N}^*\cup\{0\})=:B\space\text{and define}+\text{by}\\(a,d)+((b,g)+(c,h)):=(0,a+d+b+g+c+h) :=(a+b+c, d+g+h):=((a,d)+(b,g))+(c,h)\in B.$$
$e:=(0,0)\in B.\\$
$\text{If}\space (k,l)\in B,(k,l)+e=(0,k+l)+e=(0,k+l)=e+(0,k+l) =(k,l)\in B.$
$\therefore\space(B,+,e)\space\text{is a commutative monoid and}\space(B,+,e)\cong\space(\mathbb{N},+,0)$.
Is what I've done correct? I have an idea but it relies on me defining a monoid this way. I am not a professional and would appreciate any feedback. Thanks.