2017-01-30, 03:17 | #1 |
"Sam"
Nov 2016
2^{2}×79 Posts |
Does n have the form (a^p+-b^p)/(a+-b) for p > 2?
Is there a test to find out whether n has the form (a^p+b^p)/(a+b) or (a^p-b^p)/(a-b) for integers a and b and an exponent p > 2? Thanks for help.
Prime n are easier than ordinary n (prime or composite) since the exponents p to test are divisors of n-1. |
2017-01-30, 13:36 | #2 |
"Forget I exist"
Jul 2009
Dumbassville
10000010110001_{2} Posts |
well in the case that a is equal to b we get (2a^p)/(2a) which goes to a^(p-1) which is a powerful number regardless of the value of a in the first case and the division by 0 error in the latter case. in the cases where b>a we get things like -3(a^2)r-3a(r^2)-1 for the numerator with p=3 and -r as the denominator in the second case which ( in the example as it doesn't represent a difference of squares) is a whole value only if r divides into 1 at last check. if b<a then we get that b=a-r and we get the opposite +3(a^2)r+3a(r^2)+1 and r which once again works only if r divides 1 at last check. if you want whole numbers for the first case you have to make sure a^p+b^p divides by a+b and yes I know I'm working backwards but it may give conditions for the forms to be integer if that's what you're looking at and then we can ask is there an a-b or a+b that we can multiply to get a difference or sum of powers
Last fiddled with by science_man_88 on 2017-01-30 at 13:38 |
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