Quick Polynomials Question (1 Viewer)

[Stefano]

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?

xⁿ + yⁿ = zⁿ

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.

 

[rama_v]

What? fermats last theorem lol

Fermat's Last Theorem states that

xⁿ + yⁿ = zⁿ

has no non-zero integer solutions for x, y and z when n > 2.

http://www.pbs.org/wgbh/nova/proof/

 

[Antwan23q]

lol dickhead
try mathematical induction

 

[Stefano]

http://www.pbs.org/wgbh/nova/proof/

I went to that link and did not find the proof. Cant you just prove it on your post? It can't be that big a proof...

 

[rama_v]

I believe this is it (or part of it anyway)
http://math.stanford.edu/~lekheng/flt/wiles.pdf

Its well beyond the range of ordinary uni maths, let alone 4 unit lol
I read somewhere that only maybe 200 ppl in the world could understand it

 

[damo676767]

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(y^n-1 + x^n-1)
= xy(zⁿ) - xy(y^n-1 + x^n-1)
= xy(zⁿ - (y^n-1 + x^n-1))

i am still tying

 

[Stefano]

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.

 

[rama_v]

lol
Hilarious

 

[Stefano]

lol
Hilarious

The Bored Community has put me in charge of raising the 'HSC'ers Morale'
Just ask Slide_Rule or KFunk

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

 

[Stefano]

I was hoping justchillin could help me with this one. He likes polynomials...

 

[100percent]

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!

 

[Stefano]

rofl, this is like trying to find a general method to solving quintic equations!

It's like that, but not quite so easy

 

[damo676767]

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]

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

 

[damo676767]

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

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 x^3[/ = x³

 

[Stefano]

Haha, sorry damo. I didn't mean to waste your time. It was a joke.

 

[icycloud]

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?

xⁿ + yⁿ = zⁿ

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.

 

[Stefano]

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.

Nice try, but I didn't ask you to prove true for ALL values n>=2

 

[icycloud]

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".

 

[Stefano]

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".

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.