if p and q are two prime numbers greater than 6,then prove that p^2-q^2 is necessarily divisible by 24
Answers
Answer:
Each prime number can be shown as 6m+1 or 6m-1. for m>= 1 the prime is always bigger than 3.
If p=6m+1 and q=6n+1 then we have:
p^2-q^2=(6m+1)^2-(6n+1)^2=36m^2+12m+1-36n^2-12n+1
=36(m^2-n^2)+12(m-n)=12[3(m-n)(m+n)+(m-n)]
=12[3(m-n)(m+n+1)]
If both m and n are even or are odd then (m-n) id even ,so we have:
12(3)(2)(m+n+1) which is divisible by 24.
If m is odd and n is even or m is even and n is odd, then (m+n+1) is even and again the above term is divisible by 24.
If p=6m-1 and q=6n-1 we have:
p^2-q^2=36(m^2-n^2)-12(m-n)=12[3(m-n)(m+n-1)] and with the same way it is divisible by 24.
If p=6m+1 and q= 6n-1 then we have:
p^2-q^2=(6m+1)^2-(6n-1)^2=36m^2+12m+1-36n^2+12n-1
=36(m^2-n^2)+12(m+n)=12[3(m-n)(m+n)+(m+n)]
=12[3(m+n)(m-n+1)] and with the same way we can prove it is divisible by 24.
Answer:
Refers to the attachment
