text: simplest proof of sir fermats last theorem? please refer to the url www.in.answers.yahoo.com/ question/?qid=20060608054 655aaw70ig the question now is not if we understand your proof. the question is, do u understand what u r doing actually?  i didn't see your proof of n=a^2-b^2 for any number, but i take it as u have done it correctly becos this is true and is, in fact, very obvious. n=a^2-b^2=(a+b)(a-b)=c.d any number can be expressed as product of other numbers, this is trivial. even the nondivisible prime number can be expressed in such way by choosing a and b such that either a+b or a-b is 1.  as for your proof..  a^n = b^n + c^n  u r trying to show all numbers can only be written in a^n - b^n when n=2, and not n>2. since c^n is included in all number, hence, there is no a^n = b^n + c^n for n>2.  u started out by stating n=2. a^2-b^2. you then factor this to become (a+b)(a-b), then u restate a and b in term of x, e and d.... a=x+e, b=x+d. you then substitute a and b with x+e and x+d. then, u found for the product to be equal n, given one side is a+b, the other side must be a-b.... and hence u get back a^2-b^2. then u conclude, n must be 2 and nothing else.  the problem is.... u r essentially doing nothing here except, stating a^2-b^2... write in in another form(factor form).... write a and b in another form(x,e,d).... then u do everything in reverse again by turning (x,e,d) back into a and b.... turn factor form into a^2-b^2 form again. see. u r not doing anything at all except taking a u-turn and come back again.  when a-b=d, now matter how fancifully u express a and b, their difference is still d.  so, u dont have to repeat your proof here again and again.