Question 992626
.
Prove that {{{n^2 - 2}}} is not divisible by 5
--------------------------------------------------


For the proof let us represent the integer number  n  in the form 


n = 5k + m,  &nbsp;&nbsp;where 0 <= m < 5.


(m &nbsp;is the remainder after dividing &nbsp;n&nbsp; by &nbsp;5. &nbsp;So &nbsp;m&nbsp; can take the values &nbsp;m = 0, 1, 2, 3 &nbsp;and &nbsp;4 &nbsp;only). 


Then 


{{{n^2-2}}} = {{{(5k+m)^2 -2}}} = {{{25k^2 + 10km + m^2 - 2}}} = {{{(25k^2 + 10km)}}}) + {{{(m^2 - 2)}}} .


First two addends are divisible by &nbsp;5, &nbsp;so their sum is divisible by &nbsp;5, &nbsp;too.


Hence, &nbsp;we need to check whether &nbsp;{{{m^2 - 2}}} &nbsp;is divisible by &nbsp;5 &nbsp;for &nbsp;5 &nbsp;values of &nbsp;m = 0, 1, 2, 3 &nbsp;and &nbsp;4.


With &nbsp;m = 0 &nbsp;&nbsp;{{{0^2-2}}} = -2 &nbsp;is not divisible by &nbsp;5.


With &nbsp;m = 1 &nbsp;&nbsp;{{{1^2-2}}} = -1 &nbsp;is not divisible by &nbsp;5.


With &nbsp;m = 2 &nbsp;&nbsp;{{{2^2-2}}} = 2 &nbsp;is not divisible by &nbsp;5.


With &nbsp;m = 3 &nbsp;&nbsp;{{{3^2-2}}} = 7 &nbsp;is not divisible by &nbsp;5.


With &nbsp;m = 4 &nbsp;&nbsp;{{{4^2-2}}} = 14 &nbsp;is not divisible by &nbsp;5.


Hence, &nbsp;for any integer &nbsp;n, &nbsp;&nbsp;{{{n^2 - 2}}} &nbsp;is not divisible by &nbsp;5.


The proof is completed.