At the outset, we observe that any power of an odd number is odd and any power of an even number is even.
Then, n^7 and n must possess the same parity, and accordingly, we must have:
n^7= n (mod 2) .
n^7-n=0( mod 2), for any given integer n.......... (i)
Now, we observe that:
n^7-n =n(n^3+1)(n^3-1) .........(*)
If n=0(mod 3), then by (*), we have:
n^7-n =0(mod 3)
If n= 1(mod 3) => n^3=1(mod 3) => (n^3-1)= 0(mod 3) => n^7-n =0(mod 3)(by (*))
If n=2(mod3)=> n^3+1=0(mod 3) => n^7-n=0(mod 3) (by (*)
Accordingly, we must have:
n^7-n=0(mod 3) for any integer n. ..........(ii)
Again, as before if n=0(mod 7), then it trivially follows that:
n^7-n= 0(mod 7)
If n=1,2,4(mod 7) => n^3-1=0(mod7) => n^7-n =0(mod 7) (by *)
If n=3,5,6 (mod 7) => n^3+1 = 0(mod7) => n^7-n =0(mod 7) (by *)
Accordingly, for any given integer n, it follows that:
n^7-n=0(mod 7).............(iii)
From (i), (ii), and (iii), we observe that, for any given integer n, (n^7-n) is divisible by each of 2,3, and 7
Consequently, for any integer n, (n^7-n) must be divisible by lcm(2,3,7) = 42,
Edited on March 23, 2022, 8:32 am
Edited on March 23, 2022, 8:47 am
Edited on March 23, 2022, 8:47 am
Puzzle Solution
