Question 1070422
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Prove that for any integer n, 5 divides n^5-n.
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<pre>
{{{n^5 - n}}} = {{{n*(n^4-1)}}} = {{{n*(n^2-1)*(n^2+1)}}} = {{{n*(n-1)*(n+1)*(n^2+1)}}} = {{{(n-1)*n*(n+1)*(n^2+1)}}}.


If n is a multiple of 5, the statement is true.


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


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


If n gives the remainder 2 when divided by 5, then the factor (n^2+1) is a multiple of 5. 
     Indeed, the remainder of division by 5 is {{{2^2+1}}} = 5 (equivalent to 0) in this case, and the statement is true.


If n gives the remainder 3 when divided by 5, then the factor (n^2+1) is a multiple of 5. 
     Indeed, the remainder of division by 5 is {{{3^2+1}}} = 10 (equivalent to 0) in this case, and the statement is true.


Thus the statement is true in all cases.
</pre>

QED.  &nbsp;&nbsp;Proved and solved.