UnkleRhaukus
  • UnkleRhaukus
how come \[x^n+y^n=z^n\] only has integer solutions for n≤2
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Mani_Jha
  • Mani_Jha
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
  • JamesJ
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
  • Mani_Jha
Oh...I thought I would try this, but Andrew did it before! :P

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UnkleRhaukus
  • UnkleRhaukus
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
  • UnkleRhaukus
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
  • Mani_Jha
Well, maybe not. But I can't think of three such numbers.
UnkleRhaukus
  • UnkleRhaukus
the sum of cubes can be a square \[1^3+2^3+\dots + n^3=(1+2+ \dots+ n)^2\]
Mani_Jha
  • Mani_Jha
\[(a ^{3}+b ^{3})=(a+b)(a ^{2}-ab+b ^{2})\] 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
  • UnkleRhaukus
can a square number be a cubic number for n>2
Mani_Jha
  • Mani_Jha
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
  • UnkleRhaukus
yeah that is a perfect example
Mani_Jha
  • Mani_Jha
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.

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