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

n=0:n41=01=1n=0: n^4-1 = 0-1 = -1 is not divisible by 5.

n=1:n41=11=0n=1: n^4-1 = 1-1 = 0 is divisible by 5.

n=2:n41=161=15n=2: n^4-1 = 16-1 = 15 is divisible by 5.

n=3:n41=811=80n=3: n^4-1 = 81-1 = 80 is divisible by 5.

n=4:n41=2561=255n=4: n^4-1 = 256-1 = 255 is divisible by 5.

n=5:n41=6251=624n=5: n^4-1 = 625-1 = 624 is not divisible by 5.

n=6:n41=12961=1295n=6: n^4-1 = 1296-1 = 1295 is divisible by 5.

n=7:n41=24011=2400n=7: n^4-1 = 2401-1 = 2400 is divisible by 5.

n=8:n41=40961=4095n=8: n^4-1 = 4096-1 = 4095 is divisible by 5.

n=9:n41=65611=6560n=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,n=10a+b, a=1..9;b=0..9a=1..9; b=0..9, n is a two-digit number.


n41=(10a+b)4=(10a+b)2(10a+b)21=(100a2+20ab+b2)(100a2+20ab+b2)1=10000a4+2000a3b+100a2b2+2000a3b+400a2b2+20ab3+100a2b2+20ab3+b41.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 b41b^4-1 . We need to check all options for bb . 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 nn 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 n41n^4-1 to be divisible by 5, nn must end in any of the digits 1, 2, 3, 4, 6, 7, 8, 9.


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