I am trying to prove the following statement:
For every integer $n$, the remainder when $n^4$ is divided by $8$ is either $0$ or $1$.
So far I have figured out that $n^4 = 8m$ or $n^4 = 8m + 1$.
Any help or hints are appreciated! Thanks!
3 Answers
$\begingroup$For odd $n$, let $n=2m+1$
$\displaystyle\implies n^2=(2m+1)^2=8\frac{m(m+1)}2+1=8b+1$ where $b=\frac{m(m+1)}2$ is an integer
$\displaystyle\implies n^4=(8b+1)^2=64b^2+16b+1$
The even case is ease enough
$\endgroup$ 3 $\begingroup$Hint: $n=8k+r \to n^4=8A+r^4$ and note that $0\leq r<8$
$\endgroup$ $\begingroup$An easier way is by using the powerful tool "$mod\ 8$" for any $ n \in \mathbb{N}: n \equiv 0,1, 2, 3 ,4 , 5 , 6 \ or\ 7 \ mod\ 8 \equiv 0, 1, 2, 3, 4, -3, -2 \ or\ -1 \ mod \ 8$
so for ${n^4}$ we have
${n^4}={n^2}^2 = n^3 n \equiv 0^4, 1^4, 2^3 \times 2 , {3^2}^2, {4^2}^2, {-3^2}^2, -2^3 \times 2 \ or\ -1^4 \ mod \ 8 \equiv 0, 1, 0, 1,0,1,0,1 \ mod\ 8 $
I hope I didn't make it more complicated .
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