In my textbook I came across this problem:
Prove that the fourth power of any integer has the form $8m$ or $8m + 1$ for some integer $m$.
I know how to solve the algebra but don't understand the reasoning behind some steps of the proof.
Here is my Proof:
Suppose $n$ is an integer. By the quotient remainder theorem with $d=4$, $n$ can be written in four forms $4q$, $4q+1$, $4q+2$, or $4q+3$ for some integer $q$.
Case $1$ $(n=4q) $:
Then, $n^4=(4q)^4=256q^4=8(32q^4)$
Let $32q^4=m$. Note that $m$ is an integer since it’s the product of other integers. Hence, $n^4=8m$ for some integer $m$.
Case $2$ $(n=4q+1)$:
Then, $n^4=(4q+1)^4=256q^4+256q^3+96q^2+16q+1=8(32q^4+32q^3+12q^2+2q)+1$.
Let, $32q^4+32q^3+12q^2+2q=m$. Note that $m$ is an integer since it’s the product and sum of other integers. Hence, $n^4=8m+1$ for some integer $m$.
Cases 3 (n=4q+2):
Then, $n^4=(4q+2)^4=256q^4+512q^3+384q^2+128q+16=8(32q^4+64^3+48^2+16q+2)$
Let $32q^4+64^3+48^2+16q+2=m$. Note that $m$ is an integer since it’s the product and sum of other integers. Hence, $n^4=8m$ for some integer $m$.
Cases 4 (n=4q+3):
Then, $n^4=(4q+3)^4=256q^4+768q^3+864q^2+432q+81=8(32q^4+96q^3+108q^2+54q+10)+1$
Let $32q^4+96q^3+108q^2+54q+10=m$. Note that $m$ is an integer since it’s the product and sum of other integers. Hence, $n^4=8m+1$ for some integer $m$.
Thus, For all integers $n$, $n^4=8m$ or $n^4=8m+1$ for some integer $m$.
So, i have two main problems:
1- should i use the quotient remainder theorem?
I searched google and found that the parity property can be used instead:
2- How to know whether to use the quotient remainder theorem or the parity property in proof?