1
Rohan Ghosh
·2008-12-12 01:42:28
for the second one
let the number be even
then we get
(2n)2=4n2=4t
if it is odd
then
(2n+1)2=4n2+4n+1=4n(n+1)+1
as n(n+1) is even
=>4*2t+1=8t+1
1
ith_power
·2008-12-12 01:44:50
for 1st part only>3 primes satisfy. then 2 divides 6t+2 & 6t+4 . 3 divides 6t+3
for the second 22≡0(mod 4)
32≡1(mod 4)
1
Rohan Ghosh
·2008-12-12 01:46:49
for the first one
every prime number is odd and not divisible by 5
and p =3k+1 or 3k+2
k cannot be odd in the first case (as then it will give even)
hence p=3*(2t)+1=6t+1
and in the second case (k cannot be even)
hence
3(2m+1)+2=6m+5=6t-1
62
Lokesh Verma
·2008-12-12 01:47:43
Yup good work guys..
and good to see someone using congruence.. :)
Well for the others who did not understand the 2nd proof..
pls dont get bogged down that whole notation is not very difficult..
nor do you need to know that thing (it is simple though!)
The simple fact is that all numbers can be written as
6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5
you can easily remove some of them as a multiple of some smaller numbre!! pls complete the proof this way!
62
Lokesh Verma
·2008-12-12 01:48:20
Ooops rohan already has :)
good work :)
wasnt it the simplest question fo the day for those who did know this ;)
11
Anirudh Narayanan
·2008-12-12 02:00:11
A prime number should not be a multiple of 2 or 3. ie it should not be a multiple of 6. For a given t, numbers which are not a multiple of 6 are 6t-1, 6t+1, 6t-2, 6t+2, 6t-3 and 6t+3. But the last four are non prime. thus only 6t-1 and 6t+1 are prime.
I know this is not very convincing, but it's all i could think of. :(
11
Anirudh Narayanan
·2008-12-12 02:15:35
Sorry ith power and rohan. You guys must have made and posted ur proofs when i was making mine. Pls don't think i saw ur proofs, modified it and posted as my own (cos that's how it looks when you see the times they were posted) :)