Strong equation?

Question

a^(n) - b^(n) = (a-b)^(n) + nk(a-b)
if either "a" or "b" is odd and n is 4 or at least a prime then "k" will be an +ve natural number.
Prove that then it is impossible to write
(a-b)^(n) + nk(a-b) = 1
for n>2.

gimb a BOOM I will give
in ur first case a>b
take,a=1.25,b=0.75,n=2 & k=0.75
(1.25-0.75)^(2)+2*0.75(1.25-0.75) =1.BOOM.

gimb a BOOM I will give
in ur first case a>b
take,a=1.25,b=0.75,n=2 & k=0.75
(1.25-0.75)^(2)+2*0.75(1.25-0.75) =1.BOOM.

Answer 1

dude ..!

Answer 2

ok so K >0 n >0
1) assume a>b then BOOM not possible to write as = 1. bcouse (a-b)^(n) + nk(a-b) > 1

2) assume a
3) a=b BOOM result alsways 0. ((a-b)^(n) + nk(a-b)=0)

again you managed to do extremel;y complicated for some trivial eqaution.


Comments on the comment of the Asker :
Your BOOM is not coreect because either a or b has to be ODD, Well they are not odd and neither even

Answer 3

substitute a certain number in every variables like 2,3 or 7. then see if it is possible.