Folding a constant into the instruction
Adding a small constant is free of any load. The compiler folds the number straight into the instruction with the immediate form addi rD, rA, imm — imm being the literal value, riding along inside the opcode:
addi r3, r3, 5 # r3 = r3 + 5
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Because that immediate field is signed and 16 bits wide, the very same addi also subtracts. Want n - 3? You get addi r3, r3, -3, and not a single extra instruction. The wrinkle is how far it reaches. Sixteen signed bits cover -32768 up to 32767; ask for anything beyond and the compiler splits the work across lis plus addi. Won't happen in this exercise, but file the shape away, because it will trip you up later otherwise.
Whatever immediate the target addi carries is the constant you are after.
Your task
Write increment so it compiles to the target addi instruction.