A relation $R \subset \mathbb{N}^d$ is called recursive if there exists a primitive recursive function f with $$ (x_1 ,\dots,x_d) \in R \Leftrightarrow f(x_1,\dots,x_d)=0.$$
In Kurt Gödel's article 'Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme I', he proves the following proposition IV:
Assume $f : \mathbb{N}^n \to \mathbb{N}$ is a recursive function and $R \subset \mathbb{N}^{m+1}$ is a recursive relation. Then, the relation $S \subset \mathbb{N}^{n+m}$ $$ (a,b)\in S :\Leftrightarrow \exists x\in \mathbb{N} \text{ with }x \leq f(a) \text{ and } (x,b)\in R $$ with $a=(x_1,\dots,x_n)$ and $b=(y_1,\dots,y_m)$, is also recursive.
After this proposition he claims that it is clear that $U \subset \mathbb{N}^2$ $$ (x_1,x_2)\in U :\Leftrightarrow \exists x\in \mathbb{N} \text{ with }x \leq x_1 \text{ and } x_1 = x_2\cdot x $$ is a recursive relation. However, I do not see that this follows from proposition IV because the relation $R$ should only depend on $(x,x_2)$ not on $(x,x_1,x_2)$.
Is it still possible to show that $U$ is a recursive relation?
Because of proposition IV the relation $S$ definied by
$$ (x_1,x_2,y_1,y_2) \in S : \Leftrightarrow \exists z\in \mathbb{N} \text{ with } z \leq x_1 \text{ and } y_1 = y_2 \cdot z $$ is recursive. Therefore, it exists a recursive function $f$ with $$ (x_1,x_2,y_1,y_2) \in S \Leftrightarrow f(x_1,x_2,y_1,y_2) = 0 .$$ Since $g(x,y):=f(x,y,x,y)$ is a recursive function, we have that the relation $U$ defined by $$ (x,y) \in U :\Leftrightarrow (x,y,x,y) \in S \Leftrightarrow g(x,y)=0 $$ is recursive.