#129808 - keldon - Sun May 27, 2007 3:39 pm
http://en.wikipedia.org/wiki/Fermat's_last_theorem
How do you think Fermat could have concluded his proof for x^n + y^n = z^n has no integer solutions for n>2? The proof we have uses maths that did not exist back then. I have my suspicions at it being something to do with the following:
1: (a+1)^n - a^n is greater than a^n - (a-1)^n for any a > 0
2: a^(n+1) > a^n for any a > 1 and n > 0
3: if (a+1)^n - a^n > (a^n)*2 for any a > 1 then there is no b such that a^n + b^n >= c^n where b < c and a < c
4: there is no solution where x=1 or y=1 for any n>1
5: there is no x where x^n = y^n and x!=y for any n>0
6: for all n>2 (a+1)^n - a^n > (a^n)*2 when a=2, therefore there is no solution for any n>2
How do you think Fermat could have concluded his proof for x^n + y^n = z^n has no integer solutions for n>2? The proof we have uses maths that did not exist back then. I have my suspicions at it being something to do with the following:
1: (a+1)^n - a^n is greater than a^n - (a-1)^n for any a > 0
2: a^(n+1) > a^n for any a > 1 and n > 0
3: if (a+1)^n - a^n > (a^n)*2 for any a > 1 then there is no b such that a^n + b^n >= c^n where b < c and a < c
4: there is no solution where x=1 or y=1 for any n>1
5: there is no x where x^n = y^n and x!=y for any n>0
6: for all n>2 (a+1)^n - a^n > (a^n)*2 when a=2, therefore there is no solution for any n>2