xn is the sequence 51, 53, 57, 65, ... , 2n + 49, ... Find all n such that xn and xn+1 are each the product of just two distinct primes with the same difference.
Solution
Answer: n = 7 (x7 = 3·59, x8 = 5·61).
Calculating the first few terms we find: x1 = 3·17, x2 = 53, x3 = 3·19, x4 = 5·13, x5 = 34, x6 = 113, x7 = 3·59, x8 = 5·61.
We notice that for n odd, the term is divisible by 3. This is easily proved: 2n + 49 = (3 - 1)n + 3.16 + 1, which is a multiple of 3 for n odd. We also notice that for n divisible by 4, the term is divisible by 5. That is also easily proved: 24n + 49 = (15 + 1)n + 50 - 1, which is obviously divisible by 5.
Hence there can be no solutions with n a multiple of 4, for then 5 divides xn and 3 divides xn+1 which would imply xn = 5(p+2), xn+1 = 3p < xn. Contradiction.
If n is a multiple of 4 plus 3, then we have 3p = 2n + 49, 5(p+2) = 2n+1 + 49. Subtracting, 2p + 10 = 2n. Subtracting again, p - 10 = 49, so p = 59. It is easily checked that this is the solution n = 7, x7 = 3·59, x8 = 5·61.
n cannot be a multiple of 4 plus 2, for then xn+1 would be divisible by 3. There is no prime less than 3 except 2 and it is obvious that 2 cannot divide xn. So if xn+1 = 3p, then we would have xn = (3+k)(p+k) > xn+1. Contradiction.
Finally, we consider the case n a multiple of 4 plus 1. Then 3 divides xn, so we have 2n + 49 = 3p, 2n+1 + 49 = (3 + k)(p + k). Subtracting, 2n = 3p + kp + k2. Subtracting again, 49 = -(k-3)p - 3k - k2 (*). But we know that k is not 2, because we have just found all those solutions. Hence k ≥ 3. But that means the rhs of (*) is negative. Contradiction. So there are no solutions for n a multiple of 4 plus 2.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002