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# Let p1<p2<p3<p4 and q1<q2<q3<q4 be two sets of prime numbers such that p4−p1=8 and q4−q1=8. Suppose, p1>5 and q1>5. Prove that 30 divides p1−q1.

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Solution

## Since p4−p1=8, and no prime is even, we observe that {p1,p2,p3,p4} is a subset of {p1,p1+2,p1+4,p1+6,p1+8}Also p1 is larger than 3If p1≡1 (mod 3), then p1+2andp1+8 are divisible by 3.Hence we do not get 4 primes in the set {p1,p1+2,p1+4,p1+6,p1+8}.∴p1≡2 (mod 3) and p1+4 is not a prime We get p2=p1+2,p3=p1+6,p4=p1+8.Consider the remainders of p1,p1+2,p1+6,p1+8 when divided by 5If p1≡2 (mod 5), then p1+8 is divisible by 5 and hence is not a primeIf p1≡3 (mod 5), then p1+2 is divisibe by 5If p1≡4 (mod 5), then p1+6 is divisible by 5Hence the only possibility is p1≡1 (mod 5).Thus we see that p1≡1 (mod 2),p1≡2 (mod 3)andp1≡1 (mod 5). We conclude that p1≡11 (mod 30).Similarly q1≡11 (mod 30). It follows that 30 divides p1−q1.

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