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!
claim 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:
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.
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 .