I agree. I feel it too...

Theres a edit button you know...

+Alastair Bateman No. I think you're forgetting that x and y don't have to increase uniformly. For instance, 3,4,5 is the most well-known Pythagorean triple, but 7,24,25 also satisfies the equation, where one number is small and the other two are consecutive. The problem is WAY more difficult than just looking at how sequences of powers increase. (That's not to say that there is not a simpler proof than Ribet/Wiles, but it is certainly not as simple as you say here.)

+greg55666 The quarter square on the hypotenuse = sum of quarter squares on the other two sides. This is proof for the distributive law of multiplication. Stating the Fermat triple as (Z^n-2)(Z^2)=(Y^n-2)(Y^2)+(X^n-2)(X^2) does not obey the DLofM, a CONTRADICTION. Stating a Pythagorean triple likewise gives (Z^0)(Z^2)=(Y^0)(Y^2)+(X^0)(X^2) which works since all powers of 0=1. Q.E.D.If you derive X^n.Y^n.Z^n from the Fermat triple you get 3 other algebraic equalities from which one can easily derive 3 specific quadruples of power n. These self same quadruples are got by feeding two terms of a Pythagorean triple into either of (a+/-b)^2 and expanding the expression. This is irrefutable proof that n can only ever equal 2 for an all integer triple of the powers .Q.E.D. No known quadruples exist above the 4th power and never will not even in hyper, hyper, hyper, hyper, hyper number space.

+Alastair Bateman Hey, thanks for writing back. I would like to discuss this with you more. Could you define some terms, please? When talking about math we need to be crystal clear. What is a "quarter square"? What is "on the hypotenuse"? Why do you say something is the proof of the distributive law of multiplication? The distributive law is not a theorem; it is an axiom of the field of real numbers. Multiplication is distributive literally by definition. Why do you say factoring out an x, y, and z "does not obey the DLoM"? You haven't done anything related to the DLoM here that I can see? If you take x^4 + y^4 = z^4 and rewrite that as (x^2)(x^2) + (y^2)(y^2) = (z^2)(z^2), that is perfectly legitimate, and perfectly trivial. I don't understand what you think you are doing. What do you mean by "derive X^n.Y^n.Z^n"? What do you mean by "algebraic equalities"? What do you mean by "quadruple of power n"? I could not decipher this sentence at all: "These self same quadruples are got by feeding two terms of a Pythagorean triple into either of (a+/-b)^2 and expanding the expression." (Side comment: "irrefutable proof" is redundant. :) ) What is "hyper, hyper, hyper, hyper, hyper number space"?

+greg55666 I do not intend to get into a silly nit picking discourse via U-Tube comments with you as I can see that the comments are going to get dumb pretty fast but I will amuse you on this one occasion![1] What is a "quarter square"? Believe it or not it's a square cut into 4 quarters. 25/4=16/4+9/4 same as (5/2)^2=(4/2)^2+(3/2)^2 you know Pythagoras's theorem.[2] What's "on the hypotenuse"? Well at this moment it's 25/4 alias (5/2)^2 or a rectangle 5x1.25=6.25 sitting on the long side of a R.A.T. not the rodent variety.[N.B.] Not to be confused with 'The quarter square RULE' which is NOT a RULE but a theorem due to Euclid in Elements book 2 prop 8, brilliant in its simplicity yet profound in its outcomes. Totally ignored by the mathematical establishment yet surreptitiously resurrected as Euclid's numbers when it suits their purpose which is hypocrisy when it is just the Pythagoras equation in a slightly revised algebraic format which explains why it generates the Pythagorean triples. [3] I totally accept your comment as what I have done is, to say the least, very obtuse. However your example does not exist in the integers at least otherwise you will have proved Wiles, Fermat & myself wrong. I see and I consider so does mathematics, your example as A(Z^2)=B(Y^2)+C(X^2). For example 81^2=18^3+9^3. If we convert 729 into a cubic term we get a non integer cube root. However dividing through by 9^2 gives 9^2=18(2^2)+3^2. The cubic quadruple 6^3=5^3+4^3+3^3 also reduces to either one of two possible square triples plus a multiplier for at least one term. [4] "What do you mean by "derive X^n.Y^n.Z^n"? " : Write the Fermat equation in its 3 possible forms. Divide the sum/difference terms by the single term on the opposite side of the equation. The 3 algebraic identities all equal 1 if the triple is kosher for the integers and are therefore algebraic equalities. Multiply all three identities by X^n, Y^n & Z^n to remove the denominators. Voila!. As all 3 equations =1 then 1 becomes (X^n)(Y^n)(Z^n). If you merge the identity having sum terms with the 2 having difference terms and cancelling you get 2 quadruples of power n from which a 3rd can be derived but with one term -ve. That's a little exercise for you![5] " I could not decipher this sentence at all: " I get a feeling some ones taking the p*ss! (a+b)^2 = a^2+2ab+b^2. I'll let you do (a-b)^2 as you seem to need the practice. Now it escapes every bodies notice that (a+b)^2 is actually (a^1+b^1)^2 which as far as algebra is concerned is (a^n+b^n)^2. Again I'll let you do the expansion of the expression. With a little, not a lot, of ingenuity you will derive the self same 3 quadruples of the powers obtained in [4]. Two entirely different ways to end up with the same equations. WHAT A COINCIDENCE!!Now as I know you know the Pythagorean triple 3,4,5 then let a^n=9 & b^n=16 you will get 3 quadruples [i] 25^2=20^2+12^2+9^2 [ii] 25^2=16^2+15^2+12^2 [iii] 20^2=16^2+15^2-9^2. You will find [i] & [ii] in 'OEIS'. [iii] is not classed as a P.Q. by the establishment but I DO as it is derived from [i] & [ii]. So every Fermat triple that existed would automatically generate a Pythagorean quadruple which of course is an bsurdity so the only possible value for n is 2. ' A Rock Solid Proof'.A final comment. The 'quadrilateral rulers' n^1+n^0, triangular numbers, the squares & pronic numbers measure the powers and not vice versa so one shouldn't be surprised that the Pythagorean Theorem reigns supreme.

+Alastair Bateman There's no reason to get defensive, we're talking about math--it's all "nitpicking." I'm not arguing, I am trying to UNDERSTAND your argument. Yes, of course I "believe" a quarter square is a square cut into four quarters--why wouldn't I believe it? I'm asking for YOUR DEFINITION! Okay? I'm not attacking, I'm trying to understand; one might consider enjoying the attention (why did you write in the first place if it wasn't to be read?!) Listen, I wrote the above before I finished reading the rest of your response. I'm afraid it is more obtuse than the original. Let's stop. Sorry to bother you.