16-bit unsigned variable casting to generate correct result

Question

If all variables are uint16_t, why do you need to explicitly cast result of subtraction to yield to result?

Min test to reproduce

#include <stdio.h>
#include <stdint.h>

static void
delay_until_not_working(uint16_t start_time, uint16_t delay_time) {
    uint16_t curr_time = start_time;
    
    /* Endless loop - even if unsigned integers */
    while ((curr_time - start_time) < delay_time) {
        printf("Start: %5d, Curr: %5d, delay: %5d\r\n", (int)start_time, (int)curr_time, (int)delay_time);
        curr_time++;
    }
}

static void
delay_until_working(uint16_t start_time, uint16_t delay_time) {
    uint16_t curr_time = start_time;
    
    /* Added casting to uint16_t */
    while ((uint16_t)(curr_time - start_time) < delay_time) {
        printf("Start: %5d, Curr: %5d, delay: %5d\r\n", (int)start_time, (int)curr_time, (int)delay_time);
        curr_time++;
    }
}

int
main() {
    printf("Delay until working..\r\n");
    delay_until_working(65530, 200);
    printf("Delay until endless loop..\r\n");
    delay_until_not_working(65530, 200);
    return 0;
}

I have a 16-bit free-running timer to generate micro-seconds delay and have observed that in case of 16-bit unsigned, I need explicit cast, otherwise while loop never ends. But this works fine when variables are uint32_t. Architecture is Cortex-M with GCC 10.3.1

Question: Does subtraction operation return result in int type, when sizeof(operands) < sizeof(int)?