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Multiples of 8 plus 1
The product of n consecutive integers is always divisible by n! [i.e. n factorial, where 0! = 1, 1! = 1, 2!= 2x1, 3!=3x2x1, 4!=4x3x2x1, and so on for positive integer n.] Now let us take the odd number under test as (2n+1) for all n>=1. .............. (Theorem) So, if p = (2n+1)^2 ......................(1) then, p = 4n^2 + 4n + 1 .....................(2) => p = 4n(n+1) + 1 ............................(3) Now, as n is a positive integer greater than or equal to 1, n(n+1) is the product of two consecutive integers, which by above stated theorem, will always be divisible by 2! = 2. Thus for some positive integer k , n(n+1) = 2k ........................(4) Substituting (4) in (3), we get, p = 4n(n+1) + 1 = 4(2k) + 1 = 8k + 1 for some positive integer k. Thus, p = 8k+1 ..............................(5) Hence Proved. Q.E.D. context |