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

Question #97807
Find all positive integers n such that n^4 - 1 is divisible by 5
1
Expert's answer
2019-11-04T09:19:21-0500

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.


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