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

Question #97717
Find all positive integers n such that n^4 -1
is divisible by 5.
1
Expert's answer
2019-10-31T09:50:06-0400

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.


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