How to get a array like this?

Since you mentioned "optimal arithmetic", if you prefer doing additions over multiplications, you can take advantage of the fact that the square number of n is equal to the sum of the first n odd numbers; therefore the difference between the consecutive squares (i*i) and (i+1)*(1+1) is equal to the (i+1)st odd number (2*i+1). You can reverse that dependency to construct the square numbers like this:

void make_squares(int* square_array, int n) {
   int square = 0;                      // initialize square value
   int current_odd_number = 1;          // first odd number
   while (current_odd_number < n<<1) {  // last odd number will be 2*n-1, i. e. less than 2*n
      *square_array = square;           // assign current square value to current element
      square += current_odd_number;     // calc next square value
      current_odd_number += 2;          // calc next odd number
      ++square_array;                   // bump element pointer to next element
   }
}

Note that modern CPUs will likely be just as fast doing the multiplications as suggested in solution 1. Just wanted to point out an alternative.