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2017-04-20, 15:59   #51
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by Dr Sardonicus I have no idea of any significance to numbers of the form 10^n + 7. The polynomial x^n + 7 is irreducible in Q[x] for every positive integer n, by Eisenstein's criterion with p = 7. FWIW, I checked 10^n + 7 up to the limit n = 2000, and found pseudoprimes for the exponents n = 1, 2, 4, 8, 9, 24, 60, 110, 134, 222, 412, 700, 999, and 1383. I don't see any obvious pattern. Most (but not all) of these n are even. Six of them are divisible by 3. If n is odd, then 10*(10^n + 7) = (10^(n+1)/2)^2 + 70, so -70 is a quadratic residue of every prime factor of 10^n + 7. If one is testing small primes as divisors for the given odd value of n, this would eliminate about half the candidates.
10^(p-1)-1 is divisible by p for p coprime to 10 ( via fermat's little theorem probably an extension I'm missing to Euler's theorem.) and hence 10^(p-1)+7 is 8 mod p. and for multiples of it that fit these forms it relates to 10^(p-1)+1 etc. mod p to figure it out in this case for the exponent doubling one less than a prime it is 9 mod p. edit: okay scratch that last part my brain is wacko it seems.

Last fiddled with by science_man_88 on 2017-04-20 at 16:37