## Indirect primality t. (contd)

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• This is only to illustrate the procedure. I have considered only a) the cyclotomic polynomial x^2+1 b) values of x ranging from 1 t0 100 and c) the relevant
Message 1 of 1 , Aug 4, 2008
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This is only to illustrate the procedure. I have
considered only a) the cyclotomic

polynomial x^2+1 b) values of x ranging from 1 t0 100 and c)
the relevant failure functions

Run the following programmes in pari {p(x) = 1 + 2*k}, {p
(x) = 2 + 5*k},

{p(x) = 3 + 10*k}, {p(x) = 4 + 17*k} etc. For (x = 1,
100,print(p(x))).

Eliminate all the values of x generated by the above
programs. Now consider

only the even values of x out of the remaining. The result:

x x^2 + 1 x x^2 + 1
Remark

14 197 16 257
All the numbers in

20 401 24 577
col. x^2 + 1 are prime

26 677 36 1297

40 1601 56 3137

66 4357 74 5477

84 7057 90 8101

94 8837

Conclusion: Any large prime suspect can be indirectly
tested for primality

provided you use sophisticated programming and
test only whether

the relevant value of x is generated by the
program. If not x^2 + 1

is prime. Logic: whenever x^2 + 1 is composite
one of the factors is less

than x and hence covered by atleast one of the
failure functions.
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