SOLUTION: Prove that n^8 − n^4 is divisible by 5 for any natural n.

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Question 1033741: Prove that n^8 − n^4 is divisible by 5 for any natural n.
Found 2 solutions by robertb, ikleyn:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
N.B.: The proof given by the other tutor above is essentially the same proof here, making use of modular arithmetic, but is just much wordier and lacks further originality and insight.
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Proof:
If n is divisible by 5, then it is quite clear that n%5E8+-+n%5E4 is divisible by 5. Hence check for the other equivalence classes 5k + 1, 5k + 2, 5k + 3, 5k + 4.
It is easy to see that
%285k+%2B+1%29%5E8+=+1+%28mod+5%29
%285k+%2B+2%29%5E8+=+1+%28mod+5%29
%285k+%2B+3%29%5E8+=+1+%28mod+5%29
%285k+%2B+4%29%5E8+=+1+%28mod+5%29.
Furthermore,
%285k+%2B+1%29%5E4+=+1+%28mod+5%29
%285k+%2B+2%29%5E4+=+1+%28mod+5%29
%285k+%2B+3%29%5E4+=+1+%28mod+5%29
%285k+%2B+4%29%5E4+=+1+%28mod+5%29.
Hence,
%285k+%2B+1%29%5E8+-+%285k%2B1%29%5E4+=+0+%28mod+5%29
%285k+%2B+2%29%5E8+-+%285k%2B2%29%5E4+=+0+%28mod+5%29
%285k+%2B+3%29%5E8+-+%285k%2B3%29%5E4+=+0+%28mod+5%29
%285k+%2B+4%29%5E8+-+%285k%2B4%29%5E4+=+0+%28mod+5%29,
and the statement is proved for all natural n.






Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
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Prove that n^8 - n^4 is divisible by 5 for any natural n.
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Below is another solution to the same problem.

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Let us factor n%5E8+-+n%5E4 as far as we can:


n%5E8+-+n%5E4 = n%5E4%2A%28n%5E4-1%29 = n%5E4%2A%28n%5E2%2B1%29%2A%28n%5E2-1%29 = n%5E4%2A%28n%5E2%2B1%29%2A%28n%2B1%29%2A%28n-1%29


Now, if n is a multiple of 5, then  n%5E8+-+n%5E4  is a multiple of 5.


If n gives a remainder 1 when divided by 5, then the factor (n-1) is a multiple of 5.


If n gives a remainder 4 when divided by 5, then the factor  (n+1)  is a multiple of 5.


If n gives a remainder 2 or 3 when divided by 5, then the factor %28n%5E2%2B1%29  is a multiple of 5.


So, in any case  n%5E8+-+n%5E4  is a multiple of 5,  and the statement is proved.