Answer to Question #97807 in Combinatorics | Number Theory for Anonymous

Question

Find all positive integers n such that n^4 - 1 is divisible by 5

Answer 1

Factor the expression:

$n^4-1=(n^2-1)(n^2+1)=(n-1)(n+1)(n^2+1)=$ $=(n-1)(n+1)((n^2-4)+5)=(n-1)(n+1)((n-2)(n+2)+5)=$ $=(n-1)(n+1)(n-2)(n+2)+5(n-1)(n+1)$

The second term of the expression is divisible by 5.

By the division algorithm, any $n\in Z$ can be expressed uniquely in the form:

$n=5q+r,$

where $q\in Z$ and $r={0,1,2,3,4}$ .

If $n=5q$, then $(n-1)(n+1)(n-2)(n+2)=(5q-1)(5q+1)(5q-2)(5q+2)$ is not divisible by 5.

If $n=5q+1$, then the first term is $5q(5q+2)(5q-1)(5q+3)$, it is divisible by 5.

If $n=5q+2$, then $(5q+1)(5q+3)\cdot 5q\cdot (5q+4)$ is divisible by 5.

If $n=5q+3$, then $(5q+2)(5q+4)(5q+1)(5q+5)=5(5q+2)(5q+4)(5q+1)(q+1)$ is divisible by 5.

If $n=5q+4$, then the first term is $(5q+3)(5q+5)(5q+2)(5q+6)=5(5q+3)(q+1)(5q+2)(5q+6)$, it is divisible by 5.

Thus, $n^4-1$ is divisible by 5 when n is not divisible by 5 $(n{=}\mathllap{/\,}5k)$.

Answer: all positive integers except multiples of 5.