2012/018) For integer n if 4 is not a factor of n the prove that $5 | 1^n + 2^n + 3^n + 4^n$

4 is not a factor of n so there are 2 cases 

1) n is odd 2) n is even of the form 2(2m + 1) where m in integer.

4 is not a factor of n so there are 2 cases 

1) n is odd 2) n is even of the form 2(2m + 1) where m in integer.

Let us deal both the cases one by one

1) Case 1: n is odd

We have rearranging the terms $1^n + 2^n + 3^n + 4^n = (1^n + 4^n) + (2^n + 3^n) $

if n is odd then we have (a+b) is a factor of $a^n+ b^n$ so we get

(1+4) is a factor of $1^n + 4^n$ or 5 is a factor of $1^n + 4^n$

(2+3) is a factor of $2^n + 3^n$ or 5 is a factor of $2^n + 3^n$ 

adding the above 2 we get $5 | 1^n + 2^n + 3^n + 4^n$

2) case 2: n is even of the form 2(2m + 1)

$1^{2 * (2m+1)} + 2^{2 * ( 2* (2m + 1)}  + 3^{2 * (2m + 1)} + 4 ^ {2* (3m+1)} $

$= (1^2)^{2m+1} + (2^2)^{2m + 1}  + (3^2)^{2m + 1} + (4 ^ 2)^{3m+1} $

$= 1^{2m+1} + 4^{2m + 1}  + 9^{2m + 1} + 16^{3m+1} $

using the arguments is case 1 we know

1 + 4 or 5 is a factor of $1^{2m+1} + 4^{2m + 1}$ 

And 9 + 16 or 25 is a factor of $9^{2m + 1} + 16^{2m+1} $ and hence 5 is a factor of  $1^{2m+1} + 4^{2m + 1}  + 9^{2m + 1} + 16^{3m+1} $ or the original expression.

hence proved

as we have proved both parts so we have proved the same