Can you derive any of these results? | Moscow Math Olympiads | Divisibility

Proove: n^3 - n is divisible by 3, n^5 - n is divisible by 5, n^7 - n is divisible by 7. This video will present how to use the methods of number theory and factorization to complete all proofs.

[actions-speak]

If you prefer arithmetic to being clever, you can check that for k = 3, 5, 7 and every residue class n, n^k = n (mod k). n = 0, 1 and -1 are "freebies", and there are only 6 other cases.

[sy]

Thank you for explaining. For (a), the way is using "mod 3", and as for (c), the way is using "mod 7". So, concerning (b), why is the way using "mod 10"?
After getting n^5-n=(n-1)n(n+1)(n^2+1), if n=5k, 5k-1, or 5k+1, (n-1)n(n+1) can be divided by 5, and if n=5k±2, calculating n^2+1 is: (±2)^2+1 (mod 5) [= 0 (mod 5)]
Therefore, n^5-n is divisible by 5. (That's it. I think this method is OK and a little simpler.) (Or am I misunderstand something?)

[michaelempeigne]

one can use fermat's little theorem to prove the statements in video.

[marcgriselhubert]

It is the "little Fermat theorem: a^p is congruent to a mod p for a integer and p primitive integer. That's finished!

[ilayday]

can someone explain why its mod 10 at first?i could not understand that part.