Divisible by 24
Show that if $p > 3$ is a prime, then $24|(p^2 −1)$.
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Re: Divisible by 24
যেহেতু আমি secondary level এর তাই আমি শুধু hints দিই।
every square of an odd integer is the form of 8q+1..and primes without 2 or larger than 3, everyone are odd.again (odd^2-1)/3=kεN.(ডিভাইসিবল সাইন কিভাবে দিবো?).(8q+1-1)/3=k.And 8*3=24.
আমার solution নিয়ে আমিও confused. .............moon ভাইয়া আইদডিয়া তা ঠিক আছে???
every square of an odd integer is the form of 8q+1..and primes without 2 or larger than 3, everyone are odd.again (odd^2-1)/3=kεN.(ডিভাইসিবল সাইন কিভাবে দিবো?).(8q+1-1)/3=k.And 8*3=24.
আমার solution নিয়ে আমিও confused. .............moon ভাইয়া আইদডিয়া তা ঠিক আছে???
Re: Divisible by 24
Oh, i also secondary, bt i am giving the full solution!
Firstly we see
$p+1\not\equiv 1,3(mod 4)$ [as we khow that no prime $\ge 3$ is divisible by 2]
so, $p+1\equiv 0,2(mod 4)$
if $p+1\equiv 0(mod 4)$ then, p+1 is divisible by 4. If $p+1\equiv 2(mod 4)$ then $p-1\equiv 0(mod 4)$
so, we can say, one of p+1 & p-1 is divisible by 4. The rest one must be divisible by 2.
And then,
$p\equiv 1,2(mod 3)$
if $p\equiv 1(mod 3$ then 3|p-1. If $p\equiv 2(mod 3)$ then 3|p+1.
So, $p^2-1$ is divisible by $4\times 2\times 3=24$
\[[proved]\]
Firstly we see
$p+1\not\equiv 1,3(mod 4)$ [as we khow that no prime $\ge 3$ is divisible by 2]
so, $p+1\equiv 0,2(mod 4)$
if $p+1\equiv 0(mod 4)$ then, p+1 is divisible by 4. If $p+1\equiv 2(mod 4)$ then $p-1\equiv 0(mod 4)$
so, we can say, one of p+1 & p-1 is divisible by 4. The rest one must be divisible by 2.
And then,
$p\equiv 1,2(mod 3)$
if $p\equiv 1(mod 3$ then 3|p-1. If $p\equiv 2(mod 3)$ then 3|p+1.
So, $p^2-1$ is divisible by $4\times 2\times 3=24$
\[[proved]\]
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Re: Divisible by 24
shahriar, how did u write that odd^2 / 3 = k
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Re: Divisible by 24
just observation........Muhiminul Maruf wrote:shahriar, how did u write that odd^2 / 3 = k
Re: Divisible by 24
Press shift and \ button.AntiviruShahriar wrote:.(ডিভাইসিবল সাইন কিভাবে দিবো?)
A notable fact,we could conclude $8.3|p^2-1$ because $gcd(8,3)1$.In general if $a|c,b|c$ then $lcm(a,b)|c$
One one thing is neutral in the universe, that is $0$.
Re: Divisible by 24
Shahriar bhaia odd^2 -1 is not always dividable by 3, suppose 9^2-1 !!! but if they r prime and greater than 3 then u can write the prime number as (3n+2) or (3n+1),where n belongs to N then square it and minus 1 from it, the results will be either (9n^2+12n+3) or (9n^2+ 6n)... in both cases they are dividable by 3 @ maruf bro..
Come to the solution, notable, when p= 3n+1 then n must be even ..... then we find (p^2-1)= 3n(3n+2)...give n any even value and see n(3n+2) is dividend by 8, so 3n(3n+2) is dividable by 24, and if p= 3n+2( n here must be odd), then p^2-1= 3( 3n^2+4n +1).. give n any odd value, find 3n^2+4n+1 is dividable by 8.. so again p^2-1 is dividable by 24 if greater than 3..
Come to the solution, notable, when p= 3n+1 then n must be even ..... then we find (p^2-1)= 3n(3n+2)...give n any even value and see n(3n+2) is dividend by 8, so 3n(3n+2) is dividable by 24, and if p= 3n+2( n here must be odd), then p^2-1= 3( 3n^2+4n +1).. give n any odd value, find 3n^2+4n+1 is dividable by 8.. so again p^2-1 is dividable by 24 if greater than 3..
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