n-integr-natural-prove-that-5-divide-n-5-n-

Question Number 63645 by mathmax by abdo last updated on 06/Jul/19

n i n t e g r n a t u r a l p r o v e t h a t 5 d i v i d e n^{5} - n

Commented by Prithwish sen last updated on 06/Jul/19

n = 5 k \Rightarrow n^{5} - n divided by 5

n = 5 k + 1 \Rightarrow {(5 k + 1)}^{5} - (5 k + 1) = 5 m divided by 5

n = 5 k + 2 \Rightarrow {(5 k + 2)}^{5} - (5 k + 2) = 5 m + 30 divided by 5

n = 5 k + 3 \Rightarrow {(5 k + 3)}^{5} - (5 k + 3) = 5 m + 240 divided by 5

n = 5 k + 4 \Rightarrow {(5 k + 4)}^{5} - (5 k + 4) = 5 m + 1020 divided by 5

Hence proved .

Commented by Prithwish sen last updated on 06/Jul/19

Another method

n^{5} - n = n (n - 1) (n + 1) (n^{2} + 1)

for n = 5 k, n is divisible by 5

for n = 5 k + 1, (n - 1) is divisible by 5

for n = 5 k + 2, (n^{2} + 1) is divosible by 5

for n = 5 k + 3, (n^{2} + 1) is divisible by 5

for n = 5 k + 4, (n + 1) is divisibleby 5

∴ n^{5} - n is divisible by 5 . H ence proved .

Commented by mathmax by abdo last updated on 06/Jul/19

t h a n k y o u s i r .

Commented by Prithwish sen last updated on 07/Jul/19

welcome