how come
$$x^n+y^n=z^n$$ only has integer solutions for n≤2

Question

how come 
$$x^n+y^n=z^n$$ only has integer solutions for n≤2

mani_jha · Answer

I am yet to find a mathematical proof for it. But it seems rather intuitive, isn't it?
For n=0, it's not valid. You'll get 1=2
For n=1, it's valid. x+y=z is very true if you consider x and y to be the segments of a straight line of length z.
For n=2, it's valid. x^2+y^2=z^2 is valid if you consider x,y,z as the sides of a right-angled triangle.
After n=2, it's gonna be like:
x^3+y^3=z^3
Well, It feels rather intuitive that the cubes of two natural numbers can never add up to give the cube of another natural number. So on, for the fourth, fifth powers etc
I'll think of a proof for this. You should post this in the Maths section.
By the way, are you in college?

jamesj · Answer

Rhaukus, I think you're toying with us. This is only one of the most famous (and difficult) problems in mathematics and was solved by Andrew Wiles in 1993. If you've appetite, watch this: http://video.google.com/videoplay?docid=8269328330690408516

mani_jha · Answer

Oh...I thought I would try this, but Andrew did it before! :P

unklerhaukus · Answer

Thanks for that video i watched it all.
Still; it diden't really have any mathematics in it i want some mathematical reasoning 
... can anybody give a explanation of modular forms...

unklerhaukus · Answer

it is not intuitive in any sense to me that the cubes of two natural numbers can never add up to give the cube of another natural number

mani_jha · Answer

Well, maybe not. But I can't think of three such numbers.

unklerhaukus · Answer

the sum of cubes can be a square
$$1^3+2^3+\dots + n^3=(1+2+ \dots+ n)^2$$

mani_jha · Answer

$$(a ^{3}+b ^{3})=(a+b)(a ^{2}-ab+b ^{2})$$
Taking cube root
$$\sqrt[3]{a ^{3}+b ^{3}}=\sqrt[3]{a+b}(\sqrt[3]{a ^{2}-ab+b ^{2}})$$
cube root of (a+b) can be an integer. If we can somehow prove that the cube root of the second factor is not an integer, we might get our result.

unklerhaukus · Answer

can a square number be a cubic number for n>2

mani_jha · Answer

Oh wait if we take a=b=2, the first factor isn't an integer, but the second factor is.  
You mean like 64=4^3=8^2?

unklerhaukus · Answer

yeah that is a perfect example

mani_jha · Answer

Any number in the form of a^6 is an example of that(2^6,3^6 etc). But the cube series can never add up to a number of the form a^6.(It can be proved too). Besides we're talking of only two numbers.