Quick Polynomials Question (1 Viewer)

Stefano

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Hey guys, during my cram session I cam across this question. I know it's easy, but I have a mental block. Could someone kindly prove it?

x<sup>n</sup> + y<sup>n</sup> = z<sup>n</sup>

Prove that this equation has a non zero integer solution for n greater than 2.
Thanks. There may be a reward to whoever proves this first, because it's been bugging me for quite some time.
 
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damo676767

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when usinmg indution k =2 is easy

cause you can use and pythogorain triad to prove there is at least 1 sol

eg x=3,y=4 z=5

to prove for k = n+1 il try


lhs = xn+1 + yn+1
= x(xn + yn) + yn+1 -xyn
=xzn +y(yn + xn) -xyn - yxn
= xzn + yzn - xy(yn-1 + xn-1)
= xy(zn) - xy(yn-1 + xn-1)
= xy(zn - (yn-1 + xn-1))

i am still tying
 

Stefano

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Thanks for the proof Rama. It's so simple once you read it. Why didn't I think of that? Oh well... if they ask us that tomorrow now we know how to do it.
 

Stefano

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rama_v said:
lol
Hilarious
The Bored Community has put me in charge of raising the 'HSC'ers Morale' ;)
Just ask Slide_Rule or KFunk :D

PS. On page 109 of the proof Andrew Wiles forgot something: (Q.E.D.)
 

100percent

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Stefano said:
I was hoping justchillin could help me with this one. He likes polynomials... :)
rofl, this is like trying to find a general method to solving quintic equations!
 

damo676767

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i spent abput 2-3 hours on this question

and a massive 10+ page proof that i am not typing up proves that

x3 = x3


if any one can prove that 3n2 + 3n + 1 cannont be a perfect cube that can you tell me. it would help with this proof
 

KFunk

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Normally I would assume you're playing along with the joke but given that it's 3U-eve I'm starting to worry :p.
 

damo676767

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KFunk said:
Normally I would assume you're playing along with the joke but given that it's 3U-eve I'm starting to worry :p.

well i spend hours finding paturns, appliny formular to them

and i got out a really long equasion, and i simplified it and simplified it and proved that x3[/ = x3
 
I

icycloud

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Stefano said:
Hey guys, during my cram session I cam across this question. I know it's easy, but I have a mental block. Could someone kindly prove it?

x<sup>n</sup> + y<sup>n</sup> = z<sup>n</sup>

Prove that this equation has a non zero integer solution for n greater than or equal to 2.

Thanks. There may be a reward to whoever proves this first, because it's been bugging me for quite some time.
Disproof:

Suppose that the equation is true for n = 4:

x^4 + y^4 = z^4
(x^2)^2 + (y^2)^2 = (z^2)^2

This implies that there are solutions to Pythagoras's equation a^2 + b^2 = c^2 where a, b and c are all perfect squares. This result, of course, has been proven to be false (see Fermat's "Diophantus's Arithmetica").

The original equation x^n + y^n = z^n is false for n = 4. Therefore, it does not hold for all n >= 2 and thus is disproved by contradiction. :D
 

Stefano

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icycloud said:
Disproof:

Suppose that the equation is true for n = 4:

x^4 + y^4 = z^4
(x^2)^2 + (y^2)^2 = (z^2)^2

This implies that there are solutions to Pythagoras's equation a^2 + b^2 = c^2 where a, b and c are all perfect squares. This result, of course, has been proven to be false (see Fermat's "Diophantus's Arithmetica").

The original equation x^n + y^n = z^n is false for n = 4. Therefore, it does not hold<b> for all n >= 2 </b>and thus is disproved by contradiction. :D
Nice try, but I didn't ask you to prove true for ALL values n>=2 ;)
 
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icycloud

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Stefano said:
Nice try, but I didn't ask you to prove true for ALL values n>=2 ;)
Hehe, that's even easier then. :) The equation has an infinite number of solutions for n = 2, namely the Pythagorean Triplets. Thus, the equation "has a non zero integer solution for n greater than or equal to 2". :D
 

Stefano

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icycloud said:
Hehe, that's even easier then. :) The equation has an infinite number of solutions for n = 2, namely the Pythagorean Triplets. Thus, the equation "has a non zero integer solution for n greater than or equal to 2". :D
True!
But you seem to be missing the point my friend... I mis-phrased the question because I assumed everyone was familiar with it. You see, this is Fermat's Last Theorem - to which the solution is 109 pages of CONDENSED logic which borrows from many branches of mathematics and most of them are leading edge.

I forget the exact phrasing, but ignore the '..or equal to 2'. I'll edit my original post.
 

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