Don't confuse it with Sir fermats last theorem?

Question

Prove whether it is right or wrong
it is impossible to write
a^(n) - b^(n) = c^(2)
having natural solutions of a,b & c for n>2.
i.e. u can write
a^(2) -b^(2) =c^(2)
but u cannot write
d^(3)-f^(3)= a perfect square
or,x^(n)-v^(n) = a perfect square [sayr^(2)] when n>2 where x , v & r will have natural solution.

Answer 1

I posted this question on the Google group sci.math
and I'll give the answers they had there plus some of
my own.
Let me write the equation as x ^n -y^n = z^2, so
I can keep track of the letters. Next, either
n is a multiple of 4 or n contains an odd
prime factor p. So it suffices to consider 2 cases:
n = 4 or n = p, p an odd prime.
Further,we will assume that no 2 of x, y, z contain
a common factor.
The results I have so far are:
1. x^n - y^n = z^2 is impossible in nonzero integers
when n = 4.
Proof:
If you rewrite the equation as:

(y^2)^2 + z^2 = (x^2)^2

then you know that for some integers a and b with a > b:

y^2 = a^2 - b^2

z = 2ab

x^2 = a^2 + b^2

If you multiply the first and third equations, then you have:

a^4 - b^4 = (xy)^2

which is the same format as the original equation. Since a < x, this will give an infinite descent.

2. There are infinitely many solutions for the
equation x^3 -/+ y^3 = z^2. I will construct some of them.
Solutions to x^3 +y^3 +z^3 -3xyz = w^2
are given by
x = a^2 +2bc
y = b^2 +2ac
z = c^2 +2ab
c^2 +2ab =0 gives an infinite number of solutions, but not all.
Here are some solutions to the original equation.
1. a = -1, b - 2.
z = c^2 - 4 = 0.
c = 2 leads to 9^3 - 0^3 = 27^2
c = -2 leads to 8^3 - 7^3 = 13^2

2. a = -2, b = 9,
c^2 = 36
c= 6 leads to 112^3 + 57^3 = 1261^2.
c = -6 leads to the pretty solution
105^3 - 104 ^3 = 181^2.

Unfortunately, I have no information about
this equation for p > 3.





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Answer 2

Something yoiu might be interested in : an unverified elementary proof of Fermat by an indian prof.

http://icm2006.org/AbsDef/Shorts/abs_0403.pdf

Answer 3

If n is -ve then I think, it is possible.

Try it.