In Z mod 5, compute (a + b)5.
The Attempt at a Solution
Noticing that (a+b)5 = a5+5(a4*b+2*a3*b2+2*a2*b3+a*b4)+b5. Since 5=0 in Z mod 5, it follows that 0(a4*b+2*a3*b2+2*a2*b3+a*b4)=0 and hence (a+b)5=a5+b5.
I am just wondering if it is correct for me to say 5=0 in Z mod 5 and just substitute 0 in for 5?