16-bit and 32-bit compatibility

[natapush]

1.The expression

u16a = u16b + u16c yields a u16 bit result if int is 16 bits, this means that
u16a = u16b + u16c is compliant on a 16-bit machine.

but
u16a = u16b + u16c yields a signed 32 bit result on a 32-bit machine, so explicit type case has to be used to make the result compliant?
u16a =(u16)(u16b + u16c )

but this is not MISRA comform(rule 10.3 says type has to be narrower as the underlying type)

what is the solution for the expression u16a = u16b + u16c to be MISRA -conform on both machines?(will u16a = (u16)u16b + (u16)u16c also result in integral promotion?)

2. int16_t foo1(uint8_t x)
{
return(20000) compliant on the 16 bit machine
return(20000) not compliant on the 32 bit machine
}

right? (this is from example to rules 10.1-10.2 return(20000))