Question

Prove that 3 divides 2n^2 + 1 if and only if 3 does not divide n

Answer 1

ok so according to D.A
n=3k
or
n=3k+1
n=3k+2
we wanna prove that for n= 3k+1 and n= 3k+2

3 divides 2n^2 + 1 , and when n=3k , 3 doesn't divide it

Answer 2

Right okay, i get that so far

Answer 3

case 1:-
n=3k
2(3k)^2+1=18k^2+1=3(6k^2)+1 of the form 3m+1 which is not divisible by 3

Answer 4

case 2:-
n=3k+1
2(3k+1)^2+1=2(9k^2+6k+1)+1
=18k^2+12k+2+1
=3(6k^2+2k+1) of the form 3m which is divisible by 3

Answer 5

case 3:-
n=3k+2
2(3k+2)^2+1=2(9k^2+12k+4)+1
=18k^2+12k+8+1
=3(6k^2+4k+3) of the form 3m which is divisible by 3

Answer 6

That is a fantastic help! Thankyou very much!!

Answer 7

we are not done yet :P
its iff statement

Answer 8

now , we showed one side of the prove , conversely
let 3 divides 2n^2 + 1
2n^2+1 mod 3 =0 mod 3
2n^2 mod 3 =-1 mod 3
2n^2 mod 3= 2 mod 3

n^2=1 mod 3 , either n=1 mod 3 so n=3k+1
or
n=-1 mod 3 which means n=2 mod 3 so n=3k+2

and both are not divisible by 3 !

Answer 9

That is great help, thankyou very much :)

Answer 10

np :)