 mersenneforum.org Does n have the form (a^p+-b^p)/(a+-b) for p > 2?
 Register FAQ Search Today's Posts Mark Forums Read 2017-01-30, 03:17 #1 carpetpool   "Sam" Nov 2016 2·163 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
science_man_88

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Jul 2009
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100000110000002 Posts Quote:
 Originally Posted by carpetpool 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.
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

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