I am having troubles showing that if you could prove Fermat's Last Theorem for n = 4 and you knew it held for any odd prime then Fermat's Last Theorem holds.

I am not entirely sure where to start...

Any help would be great!

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- Jun 16th 2010, 10:02 AMRonyPonyFermat's Last Theorem
I am having troubles showing that if you could prove Fermat's Last Theorem for n = 4 and you knew it held for any odd prime then Fermat's Last Theorem holds.

I am not entirely sure where to start...

Any help would be great! - Jun 16th 2010, 10:09 AMSpringFan25
claim $\displaystyle a^n +b^n = c^n$ has no integer solutions for some values of n.

You can see that if the proposition is true for n, then it must be true for dn (where d is an integer) since:

$\displaystyle a^{dn} + b^{dn} = c^{dn} \rightarrow (a^d)^n + (b^d)^n = (c^d)^n $

So a solution for dn implies would imply a solution for n, which is a contradiction.

Its no great leap from there to show that only primes above n=4 (and n=4 itself) must be considered. - Jun 18th 2010, 01:33 AMsimplependulum
The general method to prove FLT for n = 4 is called ' method of infinite descent ' .

The process is as follows , let X be the minimum integer solution of any equation then do some analysis to show that if X exists , then we can always find another solution Y which is smaller than X , so by reductio ad absurdum we say no such X exists .

In this case, we only need to show no solution for x^4 + y^4 = z^2 only .

(x,y,z) is said to be the minimum solution of x^4 + y^4 = z^2 if z is minimum > 0 , then

x^4 + y^4 = z^2

It is fact that any two of {x,y,z} are coprime , otherwise , if two of them have a common factor d>1 , the other also has the factor d , then ( x/d , y/d , z/d ) is another solution which is smaller than (x,y,z) ,a contradiction .

Since any two of them are coprime , we can always find m,n such that

x^2 = m^2 - n^2 , y^2 = 2mn , z = m^2 + n^2 (Pythagoras' triplet )

here m and n are also coprime , and we find that from y^2 = 2mn , exactly one of m and n is even , wlog let it be n = 2N then m and N are also coprime so m = u^2 , N = v^2 , y=2uv

Also from x^2 + n^2 = m^2 we have another triplet :

x = s^2 - t^2 , n = 2N = 2st , m = s^2 + t^2

But N = v^2 = st so s = S^2 , t = T^2 . By substituting them and m=u^2 into m=s^2 + t^2 we have : u^2 = S^4 + T^4 which is in the form x^4 + y^4 = z^2 , so we say if there exists such triplet (x,y,z) we can also find another one ( S,T ,u ) . However , z = m^2 + n^2 > m^2 = u^4 > u . By reductio ad absurdum , we say no such triplet exists .