text: The question now is not if we understand your proof. The question is, do u understand what u r doing actually?\n\nI 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.\n\nAs for your proof..\n\na^n = b^n + c^n\n\nu 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.\n\nU 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.\n\nThe 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.\n\nwhen a-b=d, now matter how fancifully u express a and b, their difference is still d.\n\nSo, u dont have to repeat your proof here again and again.