Thread: Always an integer. View Single Post
2007-07-12, 08:33   #11
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
General Eqn.

Quote:
 Originally Posted by davieddy Neat. Now we can generalize the problem to contain any number of terms x^p/p where p is prime. e.g. x^7/7 +x^2/2 + 5x/14 David

Excellent Davie!

I would like you to note that the general expression is true for all +ve integers when p is a prime It is also true if p divides x. This is a corollary which results from Fermat’s little theorem.

Thus from your very expression when x is any number e.g. x = 14
14^7/7 + 14^2/2 +5*14/7 an integer although 7 and 2 both divide 14

A more restricted expression can be derived from Fermat’s little theorem, thus
x^6/7 + x^2/3 – 10/21 is also true and always an integer provided the primes
(In denominator) do not divide x , and the constant is changed accordingly as it’s not a function of x but rather of the primes.

Having said that could you derive the general expression in terms of x, p_1, and p_2……?

Try it with x = 5, 11 and any prime.

Mally .