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## A theorem about prime numbers

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• 1 * 1 = 7 * 13 = 11 * 11 = 17 * 23 = 19 * 19 = 29 * 29 = 1 mod 30 For the integer k, If for every value of n, none of m1 = (k-n)/(30*n+1) m2 = ( k - 7 * n -
Message 1 of 1 , Apr 9 6:42 PM
1 * 1 = 7 * 13 = 11 * 11 = 17 * 23 = 19 * 19 = 29 * 29 = 1 mod 30

For the integer k,
If for every value of n, none of

m1 = (k-n)/(30*n+1)
m2 = ( k - 7 * n - 3) / (30 * n + 13)
m3 = (k - 11 * n - 4)/(30 * n + 11)
m4 = (k - 13 * n - 3)/(30 * n + 7)
m5 = (k - 17 * n - 13) / (30 * n + 23)
m6 = (k - 19 * n - 12) / (30 * n + 19)
m7 = (k - 23 * n - 13) / (30 * n + 17)
m8 = (k - 29 * n - 28) / (30 * n + 29)

are an integer,
then
(30 * k + 1) is a prime.

Derivation:

(30 m + 1) * (30 n + 1) = (30 k + 1)
30 * 30 * m * n + 30 n + 30 m + 1 = 30 k + 1

k = 30 * m * n + m + n

30 * m * n + m = k - n

m = (k - n) / (30 * n + 1)

Maximum n is found by:

n = (k - n) / (30 * n + 1)

30 * n^2 + n = k - n

30 * n^2 + 2 n - k = 0

max n = - 1 + sqrt(1 + 30 k)

(30 m + 7) * (30 n + 13) = (30 k + 1)

30 * 30 * m * n + 7 * 30 n + 13 * 30 m + 91 = 30 k + 1

k = 30 * m * n + 7 * n + 13 * m + 3

30 * m * n + 13 * m = k - 7 * n - 3

m = ( k - 7 * n - 3) / (30 * n + 13)

max n is found by

n * (30 * n + 13) = (k - 7 * n - 3)

30 * n^2 + 13 n = k - 7 * n - 3

30 * n^2 + 20 n - k + 3 = 0

max n = -10 + sqrt(100 + 30 * (k-3) )

etc
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