1as nishant said evry prime no above 5 can be represented as 6k±1, so p2-1=(6k+1)2-1
=36k2+12k+1-1
=12(3k+1)
so it is divisible by 12.................but y 24????
Is it true that (p2-1) is always divisible by 24 if p is a prime number greater than equal to 5 ? if Yes how ?
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12 Answers
1dats true.. coz all prime numbers greater than or equal to 5 on squarin are divisible by the multiples of 8...
341every prime number is of the form 4k±1
So, of p+1 and p-1, one will be a multiple of 4 and the other of 2.
Hence p2-1 is divisible by 8
Now, consider the three consecutive numbers p-1, p and p+1
One of the has to be divisible by 3 and it obviously cannot be p (as p is prime)
So, one among p-1 and p+1 is a multiple of 3 ensuring that p2-1 is also a multiple of 3
Since gcd(3,8) is 1, p2-1 is a multiple of 24
1
62think like this
36k2 ± 12k
= 24k2 +12k2 ± 12k
Now does it become more obvious!?
*** Edit to complete the proof***
= 24k2 +12k( k ± 1 )
if k is odd ( k ± 1 ) is even
if k is even ( k ± 1 ) is odd..
hence k( k ± 1 ) is alway even
so the last term is also divisible by 24 :)
** Ended Complete proof*****
1both the approaches r correct...
try this 1...
n7 - n is divisible by 42.
1these questions can be very easily solved using Fermat's Little Theorem which states that ap-a is always divisible by p provided p os a prime
now in this case n7-n will be divisible by 7 now it can be factorized like
{n(n-1)(n+1)(n2+1-n)(n2+1+n)
now since above expression is divisible by 7 and
n(n-1) is divisible by 2 and n(n-1)(n+1) is divisible by 3
hence we can say that above complete expression is divisible by 7*3*2=42
1these questions can be very easily solved using Fermat's Little Theorem which states that ap-a is always divisible by p provided p os a prime
now in this case n7-n will be divisible by 7 now it can be factorized like
{n(n-1)(n+1)(n2+1-n)(n2+1+n)
now since above expression is divisible by 7 and
n(n-1) is divisible by 2 and n(n-1)(n+1) is divisible by 3
hence we can say that above complete expression is divisible by 7*3*2=42
11Couldn't we do the same thing for the 1st one?
Odd squares are of the form 8k+1, meaning p2-1 is divisible by 8.
FLT gives p2-1 ≡0(mod3).
Finished.