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性質：5以上的質數除以6不是餘1，就是餘5

p^2-q^2=(p+q)(p-q)情況1：p,q除以6皆餘1

令p=6m+1，q=6n+1，m,n為整數，則

(p+q)(p-q)=(6m+6n+2)(6m-6n)=12*(3m+3n+1)(m-n)

(3m+3n+1)與(m-n)恰有一個是偶數，

所以(p+q)(p-q)=12*(3m+3n+1)(m-n)為24的倍數情況2：p,q除以6皆餘5

令p=6m+5，q=6n+5，m,n為整數，則

(p+q)(p-q)=(6m+6n+10)(6m-6n)=12*(3m+3n+5)(m-n)

(3m+3n+5)與(m-n)恰有一個是偶數，

所以(p+q)(p-q)=12*(3m+3n+5)(m-n)為24的倍數情況3：p除以6餘1，q除以6餘5

令p=6m+1，q=6n+5，m,n為整數，則

(p+q)(p-q)=(6m+6n+6)(6m-6n-4)=12*(m+n+1)(3m-3n-2)

(m+n+1)與(3m-3n-2)恰有一個是偶數，

所以(p+q)(p-q)=12*(m+n+1)(3m-3n-2)為24的倍數情況4：p除以6餘5，q除以6餘1

令p=6m+5，q=6n+1，m,n為整數，則

(p+q)(p-q)=(6m+6n+6)(6m-6n+4)=12*(m+n+1)(3m-3n+2)

(m+n+1)與(3m-3n+2)恰有一個是偶數，

所以(p+q)(p-q)=12*(m+n+1)(3m-3n+2)為24的倍數所以(p^2-q^2)恆為24的倍數。

p2-q2=(p+q)(p-q)*(1)* we want toprove 8|p2-q2Let p=2m+1 and q=2n+1.we know that2|(p+q) and 2|(p-q)so p-q=4k or p-q=4k+2 if p-q=4k,then 8|(p+q)(p-q) or 8| p2-q2if p-q=4k+2, then p+q=4k+2+2q=4k+2+2(2n+1)=4k+4n+4=4(k+n+1)so 4|(p+q),and 8|(p+q)(p-q) or8| p2-q2.*(2)* we want toprove 3|p2-q2For p and q aretwo primes bigger than 5,so we can assume thatp=1 or 2(mod 3) and q=1 or 2(mod 3),sincep and q can not be 3 and be divided by 3.If p=q=1(OR 2),p-q=0(mod3).If p≠q,p+q=1+2=0(mod 3).Hence, one of p+qand p-q is equal to 0 (mod 3), i.e. 3|(p+q) or 3|(p-q), implies 3|(p+q)(p-q).For 24=3*8,24|(p+q)(p-q) or 24| p2-q2 by (1) and (2).Q.E.D. assume