prove the ultimate.?

Question

prove that a^(n)+b^(n)+........+n^(n) can be expressed as difference of at least one set of two perfect squares. here n is upto infinity.

here infinity means upto number which you can think not the concept of infinity.

i am interested in natural and rational numbers .

as i already told that the number 2 and the numbers having a single 2 as its multiple cannot be expressed as difference of two perfect squares.

Answer 1

Not all a^n +b^n +c^n +... n^n=(odd)x(odd) or (even)x(even) !!!

here!

1^3 +2^3 +3^3 = 794 = 2 x 397 which is (even) x (odd)

so, 1^3+2^3+3^3 cannot be expressed as the difference between 2 perfect squares!!



concerning your latest comment,

u do not have to tell me because I can even explain to u why a single factor 2 cannot work.


Any number a^n+b^n+..+n^n is a number. It can EITHER be expressed as x^2-y^2 OR it cannot be. There's only 2 possibilities!! By saying it has a single factor 2, it simply means it cannot be expressed as x^2-y^2. so, u have simply omitted all those that cannot be expressed that way. so, those left are of course expressible as x^2-y^2, for integer x and y. This is an empty statement!! It's like saying, all number xy+z is odd EXCEPT when it is even.... it's obvious, trivial and useless.

Answer 2

well there is a problem with the questions, if n->infinity then n^(n) at infinity would be infinity^(infinity) which = inifinity

Answer 3

if n=π your theorem is untrue.