Question

Prove for every integer n, n4 has the form 8m or 8m + 1 for some integer m. Hint: Use 2 cases.

Answer 1

Let n is an even number

Then for some integer m, we can write

$n = 2m$

$n^{4}=(2m)^{4}$

$n^{4}=16m^{4}$

$n^{4}=8(2m^4)$

$n^{4}=8p$ here $p=2m^4$ an other integer

Let us consider n is an odd integer,

Then for some integer m, we can write,

$n = 2m +1$

Then

$n^{4} = (2m +1)^{4}$

Using binomial expansion,

$n^{4} = 16m^4+32m^3+24m^2+8m+1$

$n^{4} = 8(2m^4+4m^3+3m^2+m)+1$

$n^{4} = 8q+1$ here $q=2m^4+4m^3+3m^2+m$

Hence proved that

For every integer $n$, $n^4$  has the form $8m$  or $8m + 1$ for some integer m.