Answer to Question #97717 in Combinatorics | Number Theory for akash kumar

Question

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

Answer 1

1) If $n=b, b=0..9$ is one-digit number. We need to check all of them.

$n=0: n^4-1 = 0-1 = -1$ is not divisible by 5.

$n=1: n^4-1 = 1-1 = 0$ is divisible by 5.

$n=2: n^4-1 = 16-1 = 15$ is divisible by 5.

$n=3: n^4-1 = 81-1 = 80$ is divisible by 5.

$n=4: n^4-1 = 256-1 = 255$ is divisible by 5.

$n=5: n^4-1 = 625-1 = 624$ is not divisible by 5.

$n=6: n^4-1 = 1296-1 = 1295$ is divisible by 5.

$n=7: n^4-1 = 2401-1 = 2400$ is divisible by 5.

$n=8: n^4-1 = 4096-1 = 4095$ is divisible by 5.

$n=9: n^4-1 = 6561-1 = 6560$ is divisible by 5.

Among numbers from n=1 to n=9, any number n except 0 and 5 will satisfy the required property.

2) If $n=10a+b,$ $a=1..9; b=0..9$, n is a two-digit number.

$n^4-1=(10a+b)^4 =(10a+b)^2(10a+b)^2-1=(100a^2+20ab+b^2)(100a^2+20ab+b^2) -1= 10000a^4+2000a^3b+100a^2b^2+2000a^3b+400a^2b^2+20ab^3+100a^2b^2+20ab^3+b^4 -1.$

Each term in this sum is a multiple of 5 except $b^4-1$ . We need to check all options for $b$ . But this is the same as it was done in part 1).

Among numbers from b=1 to b=9, any number b except 0 and 5 will satisfy the required property. It means that the last digit of the number $n$ can be any number except 0 and 5.

3) Similarly, one can show the same conclusion for a three-digit number and more.

Answer: in order for $n^4-1$ to be divisible by 5, $n$ must end in any of the digits 1, 2, 3, 4, 6, 7, 8, 9.