# n^3+n=m^4 proving n has to be even and divisble by 16

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• Sep 22nd 2011, 11:35 AM
jjy0127
n^3+n=m^4 proving n has to be even and divisble by 16
How do you prove that n^3+n=m^4 n has to be even and also divisble by 16

Thanks
• Sep 22nd 2011, 11:59 AM
anonimnystefy
Re: n^3+n=m^4 proving n has to be even and divisble by 16
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?
• Sep 22nd 2011, 02:45 PM
jjy0127
Re: n^3+n=m^4 proving n has to be even and divisble by 16
Thank you for your reply, it seems as though question is asking n has to be even not n^3+n is even.