try this and earn ten points.?

Question

1).a^(n)-b^(n)can be expressed as
difference of at least one set of two squares
like 7 ^ (3) - 4 ^ (3) = 48 ^ (2) - 45 ^ (2)
2).any number say c having any power n can be
expressed as difference of at least one set of
two squares like
7 ^ (3) = 172 ^ (2) - 171 ^ (2)
from this prove that
a ^ (n) - b ^ (n) = c ^ (n)
is not possible.
hint :- (any number) ^ (n) cannot be 0 for any value of n.

gimb i think u didnot read my statement that any number having any power can be expressed as difference of at least one set of two perfect squares and the power includes 1 and 2.

try to express 5^(7)-9^(7) into difference of two squares gimb.

sorry it is 9^(7)-5^(7)

1=1.25^(2)-0.75^(2)

Answer 1

I have found a simple and elegant proof for this, but the area that Yahoo makes available for answers is too small to contain it.

Answer 2

why are you posting the same question again ?
I will never answer a question of you again.


gjmb1960
1 hour ago



the prob is a bit ill

NB x^0 for x!=0 is by definition 1.

for n = 1 the statement is false.
a - b = x^2 - y^2
c = x^2 - y^2

for instance take a = 17, b = 1, c = 16, 16 = 4^2 = 5^2 - 3^2

for n = 2 the statement is false
a = 5, b = 3, c = 4, x = 5,y=3

for n>2 is true because of fermat theoreom prved by wiles.


BTW
Did you notice that the "hidden thing" in

7 ^ (3) - 4 ^ (3) = 48 ^ (2) - 45 ^ (2)

is 3^0 ?


"gimb i think u didnot read my statement that any number having any power can be expressed as difference of at least one set of two perfect squares and the power includes 1 and 2."

It is simply not true try to write 1 as difference of 2 squares
try 1^1 = x^2 - y^2 = (x-y)(x+y)
thus x-y = 1 and x+y = 1 OR x-1=-1 and x+y=-1 doesnt make sense.
------------------------------
ok you alllow quotients for the numbers,

for n = 1 the statement is false.
take a = 17, b = 1, c = 16

then 17^1 - 1^1 = 5^2 - 3^2
16^1 = 5^2 - 3^2
and 17^1 - 1^1 = 16^1 BOOM. { a ^ (n) - b ^ (n) = c ^ (n) is not possible }

for n = 2 take
a = 5, b = 3, c = 4
then
a^2 - b^2 = a^2 - b^2; can be written as difference of two squares
c^2 = a^2 - b^2; can be written as difference of two squares

but
a^2 - b^2 = c^2 ; BOOM { a ^ (n) - b ^ (n) = c ^ (n) is not possible }

for n>=3 suppose a^n - b^n = c^n then wiles was wrong BOOM.

Answer 3

4

Answer 4

You are essentially asking us to prove Fermat's last theorem, which states that the Diophantine equation:
a^(n) = c^(n)+b^(n)
has no solutions for n>2. This was proved by Andrew Wiles in 1995, and the given proof is highly non-elementary. If you have a simpler proof, I would love to see it, and so would the rest of the world.

Answer 5

42

Answer 6

52

Answer 7

69

Answer 8

The ansewr is 1.