Problem: The largest of the following integers which divides each of the members of the sequence 15−1,25−2,35−3,…,n5−n,… is:
Answer Choices:
A. 1
B. 60
C. 15
D. 120
E. 30
Solution:
Since n5−n=(n−1)n(n+1)(n2+1)=k(n2+1), where k is a product of 3 consecutive integers, k is always divisible by 6 at least. We show next that in addition, n5−n is always divisible by 5. If we divide n by 5, we get a remainder whose value is either 0,1,2,3, or 4; i.e.,
n=5q+r,r=0,1,2,3 or 4.
By the binomial theorem,
n5−n=(5q+r)5−(5q+r)=(5q)5+5(5q)4r+10(5q)2r2+10(5q)2r3+5(5q)r4+r5−5q−r.=5N+r5−r,where N is an integer If r=0,n5−n=5N is divisible by 5. If r=1,n5−n=5N is divisible by 5. If r=2,n5−n=5N+30=5(N+6)is divisible by 5. If r=3,n5−n=5N+240=5(N+48)is divisible by 5. If r=4,n5−n=5N+1020=5(N+204) is divisible by 5. Thus, for all integers n,n5−n is divisible by 5 and by 6 hence by 30;
or
n5−n=n(n4−1) is divisible by 5 since np−1≡1(modp) where p is a prime. ∴n5−n is divisible by 30.