How do you prove that n^3+n=m^4 n has to be even and also divisble by 16
Thanks
hi jjy0127
let's look at the parity of the expression n^3+n
1. case-n is even:
when n is even,n^3 is also even,so n^3+n is even.
2. case-n is odd
when n is odd,n^3 is also odd,so n^3+n is even (because a sum of two odd numbers is an even number)
so we have that the right side is always even and divisible by 2.
now,because the left side is divisible by 2 so must be the right side.so:
2|m^4
which is equivalent to:
2|m
so if m is divisible by two you can write it like this:
m=2k where k is a natural number.
from here you get that:
m^4=(2^4) * (k^4)=16*(k^4)
which means that m is divisible by 16
can you continue?