Let-P-n-n-1-n-3-n-5-n-7-n-9-What-is-the-largest-integer-that-is-a-divisor-of-P-n-for-all-positive-even-integers-n-

Question Number 19403 by Tinkutara last updated on 10/Aug/17

Let P (n) = (n + 1) (n + 3) (n + 5) (n + 7) (n + 9) .

What is the largest integer that is a

divisor of P (n) for all positive even

integers n ?

Commented by RasheedSindhi last updated on 12/Aug/17

15

Answered by RasheedSindhi last updated on 12/Aug/17

P (n) = (n + 1) (n + 3) (n + 5) (n + 7) (n + 9)

n \in E^{+}; Let n = 2 m, m \in N

P (2 m) = (2 m + 1) (2 m + 3) (2 m + 5) (2 m + 7) (2 m + 9)

m = 3 k ∣ m = 3 k + 1 ∣ m = 3 k + 2

C - 0 : m = 3 k \Rightarrow

P (n) = P (2 m) = P (2 . 3 k)

= (6 k + 1) (6 k + 3) (6 k + 5) (6 k + 7) (6 k + 9)

= 3 (6 k + 1) (2 k + 1) (6 k + 5) (6 k + 7) (6 k + 9)

C - 1 : m = 3 k + 1 \Rightarrow

P (n) = (6 k + 3) (6 k + 5) (6 k + 7) (6 k + 9) (6 k + 11)

= 9 (2 k + 1) (6 k + 5) (6 k + 7) (2 k + 3) (6 k + 11)

C - 2 : m = 3 k + 2 \Rightarrow

P (n) = (6 k + 5) (6 k + 7) (6 k + 9) (6 k + 11) (6 k + 13)

= 3 (6 k + 5) (6 k + 7) (2 k + 3) (6 k + 11) (6 k + 13)

In all the three cases :

3 ∣ P (n)

Similarly we can prove that

5 ∣ P (n)

Also can be verified from P (10)

that 3 and 5 can divide P (n) once only .

P (10) = 11.13.15.17.19

= 11.13.3.5.17.19

From the following it is proved

that there {isn}^{'} t any divisor of P (n)

other than 3 & 5

P (2) = 3.5.7.9.11

P (10) = 11.13.15.17.19

P (12) = 13.15.17.19.21

Hence the largest divisor is

3 \times 5 = 15

Commented by Tinkutara last updated on 12/Aug/17

Thank you very much Sir!