Is it possible without using any array to assign value of an integer to a designated address?

I'm recently reading about pointers in C but stuck at a point. I need to provide an example to make the question more clear.

I have declared two integer variables as a and b and assigned values 55 and 88 to them. Now by using the & symbol I can access their address and declare pointers as follows:

#include <stdio.h>

int main(void)
{
    int a = 55;
    int b = 88;

    int *pa = &a;
    int *pb = &b;

    printf("Address of 'a' is: %p\n", &a);
    printf("Address of 'b' is: %p\n", &b);

    printf("Desired new address for the value of 'b' is: %p\n", (pa+1));

    //What to do here to assign the value of b into the address (pa+1) ?
}

What I try to achieve(without using any arrays) is that: I want the value of b to be assigned to the address next to a.

I can obtain the address of a by &a or *pa = &a and the next address would be (pa+1). Now the pointer pa holds the address to the variable a. And so the value of (pa+1) I guess should point the address of the next possible address of a.

Now my problem is having the value of b and the address (pa+1) how can we register/assign the value of b to the address (pa+1)?

edit:

Following doesn't work:

int a = 55;
int b = 88;

int *pa = &a;
int *pb = &b;

printf("Address of 'a' is: %p\n", &a);
printf("Address of 'b' is: %p\n", &b);

printf("Desired new address for the value of 'b' is: %p\n", (pa+1));

*(pa + 1) = b;

printf("New address of 'b' is now: %p\n", &b);

Outputs:

Solved:

    int main(void)
    {
        int a = 55;
        int b = 88;
        int c;

        int *pa = &a;
        int *pb = &b;

        printf("Address of 'a' is: %p\n", &a);
        printf("Address of 'b' is: %p\n", &b);

        printf("Desired new address for the value of 'b' is: %p\n", (pa+1));

        *(pa + 1) = b;

       printf("New value at pa+1 is now: %d\n", *(pa+1));

}