Re: a prime number question

I was thinking of Fermat's Little Theorem: a^(p-1) = 1 (mod p). So I
thought of a prime p such that p-1 was divisible by 7. I got lucky with 29.


________________________________
Eric J. Wingler (wingler@xxxxxxxxxxxx)
Dept. of Mathematics and Statistics
Youngstown State University
One University Plaza
Youngstown, OH 44555-0001
330-941-1817

"Robert Israel" <israel@xxxxxxxxxxx> wrote in message
news:dr5q9r$sv3$1@xxxxxxxxxxxxxxxxxxxxxxxxx
> In article <43d647b8$1@xxxxxxxxxxxx>,
> Eric J. Wingler <wingler@xxxxxxxxxxxx> wrote:
> >Calculating mod 29, you get
> > 3^7 - 2^7 = 3*(3^3)^2 - 2^7
> > = 3*2^2 - 2^7
> > = 4*(3 - 2^5)
> > = 0 (mod 29).
> >Hence 3^7 - 2^7 is divisible by 29.
>
> But how did you think of using 29?
>
> BTW, there seem to be a lot of primes of the form (x+1)^7 - x^7:
> this is prime for 161 integers from 1 to 1000. In fact, the only
> primes < 100 that can divide (x+1)^7 - x^7 are 29, 43 and 71.
>
> Robert Israel israel@xxxxxxxxxxx
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada
>
>


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