tag:blogger.com,1999:blog-12535639.post111638650961365927..comments2018-03-17T17:58:54.013-07:00Comments on Fermat's Last Theorem: Fermat's Last Theorem: n = 4Larry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger63125tag:blogger.com,1999:blog-12535639.post-55399386945012451602016-12-06T13:00:21.909-08:002016-12-06T13:00:21.909-08:00Here is a simple proof
To recap, Michael has agr...Here is a simple proof<br /><br /><br />To recap, Michael has agreed that if n is an even number, particularly when n is 4, there4 exists a second Pythagorean triple with h^2,x^2, y^2 where h^2 = x^2 + y^2 <br /><br />Because this is a Pythagorean triple, there exist some whole numbers r,s such that y = 2rs, and say s is even. Squaring both sides y^2 = 4 r^2 y^2. <br /><br />Using Euclid's formula for the larger Pythagorean triple set, z^2, x^2, y^2 <br /><br />x^2 = h^2 - y^2 <br /><br />or substituting r,s for h,y and collecting terms <br /><br />x^2 = 4r^2s^2 - r^2 <br /><br />or <br /><br />x = ( 4 r^2 s^2 - r^2 )^1/2 <br /><br />or <br /><br />x = r(4s^2 - 1)^1/2 <br /><br />since s is even, values of s are 2,4, 6, 8, etc. <br /><br />x must always be a irrational number.<br />This is one method. <br /><br />After I made the last post, I realized that even though you could not extend Fermat's proof for n=4, you could use it as it is to prove the case for n being all the other even numbers higher than 2. <br /><br />Here is what I said previously. <br /><br />Once the case for n =4 is established it is rather easy to see how to extend the proof for the n being even above 2. And the reason is rather simple,since x^8 is just some other number j, where j = x^2. <br /><br />So now you have 23 different proofs of FLT for the evens. <br /><br />You tell me which one proof you agree with and I will demonstrate the case with then use that agreement to demonstrate the csase for n being odd.<br />A Response by Kevin :<br />2016-11-24 at 09:39GMT<br />Ok, <br /><br />And how do you prove other cases (n=3,5,7,9,...) ? <br /><br />What you're showing is already well known (only the way is funny, the result is obvious since two centuries).<br /><br />Thread:<br /><br /> <br />Here is the equation from the funny proof.<br /><br />x = r(4s^2 - 1)^1/2 <br /><br />From this equation you know that the part that determines whether x is rational or irrational is (4s^2 - 1)^1/2 . Let us call that part k.<br /><br />Therefore k = (4s^2 - 1)^1/2 <br /><br />So x^h = r^h k^h where h is some odd number.<br /><br />While we cannot say anything about this number when h is odd, we can say something about x when we are talking about x^(h-1)<br /><br />x^(h+1) ) is an irrational number from the proof that FLT when n=4. <br /><br />x <br /><br />x = r(4s^2 - 1)^1/2 <br /><br />the part of x that is irrational is <br /><br />(4s^2 - 1)^1/2 <br /><br />Looking at this portion , we can say something about the position on the number line. <br /><br />If we say that we look at some arbitrary odd number, j, we can look at the 2 values of x when the power of of n is j-1 and j+1. We know that x is irrational when power of n is even. <br />Let us call the irrational part of x when n = j+1, h. And let us call the irrational part of x when n = m. <br /><br />So now if we look at the number line between (4s^2)1/2 and (4s^-2)^2. Placing h,x, and m on this number line looks like this <br /><br />---(4s^2)^1/2 -1------- (4s^2 - 2)^1/2 ------ m ------- x -------- h -------- (4s^2)^1/2 ---- (4s^2)^1/2<br /><br />x is sandwiched between 2 irrational numbers that are sandwiched between 2 consecutive whole numbers. There fore x is irrational<br /><br />to see the full argument on the fake math site go to<br />http://math2.org/mmb/thread/44919<br />Reliable Home Inspectionhttps://www.blogger.com/profile/09397205998512916607noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-85421427759114845552016-08-25T14:54:48.219-07:002016-08-25T14:54:48.219-07:00Anyone who can devise a simple proof of Fermat'...Anyone who can devise a simple proof of Fermat's Last Theorem and publishes it on the Unsolved Problems web site at http://unsolvedproblems.org/ is eligible for a prize of US$2500 plus a bottle of champagne! Unknownhttps://www.blogger.com/profile/06030021311779507460noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-18978736795341609632016-07-11T11:12:14.647-07:002016-07-11T11:12:14.647-07:00@JulieL, I believe it works the other way. If the...@JulieL, I believe it works the other way. If there is no solution to x^4 - y^4 = z^2, then, there is no solution to x^4 + y^4 = z^4.<br /><br />If z is not a square, then it is possible (logically) that there is a solution to x^4 - y^4 = z^2 but no solution to x^4 + y^4 = z^4.Larry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-35504157882052083032016-07-10T09:59:51.182-07:002016-07-10T09:59:51.182-07:00Could you please help me understand why if there i...Could you please help me understand why if there is not a solution to x^4 + y^4 =z^4, then there is no solution to x^4 - y^4 =z^2 ?JulieLhttps://www.blogger.com/profile/06093306823041945080noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-82488859120903616852015-02-05T20:18:06.000-08:002015-02-05T20:18:06.000-08:00Larry, I believe that your infinite descent argume...Larry, I believe that your infinite descent argument may be flawed, as stated (or at the very least, incomplete). The flaw, as I see it, is that in step 5, when you combine equations, you are able to compare to x2, which is obviously a square; however, when you try to repeat this argument in step 8, you would only be able to compare to q, which CANNOT be a square, because of the factor of 2 in q=2ab, and that a and b are squares, so q/2 is a square, hence q is not(also, we know p is a square and so is x2, so from x2=2pq, 2q must be a square, so again q cannot be). I can only assume that the "same properties" you are referring to in step 9 is that you have come across another Pythagorean triple (P2=a2+b2) which is smaller than (y2+q2=p2). However, it does not indeed possess the same properties, as you suggest.<br /><br />However, as Scouse Rob pointed out, since ab is a square and a and b are relatively prime, a and b are also squares, hence the infinite descent can quickly be salvaged, as follows:<br /><br />(6) Since ab and a2+b2 are relatively prime, we know that they are both squares. <b>Further, since ab is a square and a and b are relatively prime, both a and b are squares as well.</b><br /><br />(7) So there exists <b>P, A, and B such that P2=p and A2=a and B2=b, respectively.</b><br /><br />(8)Now we have reached infinite descent since: <b>P2=A4+B4</b>...<br /><br />-SeanS. G. Saundershttps://www.blogger.com/profile/09671222131180895268noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-64301345287242293152014-06-01T18:03:58.753-07:002014-06-01T18:03:58.753-07:00I don't know what school of Mathematical logic...I don't know what school of Mathematical logic you attended but your argument falls down on the last line, ie D implied C so D had no solutions is nonsense as the implication is the wrong way.<br />C would have to imply D to prove that.Jezhttps://www.blogger.com/profile/03222601835879357192noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-18289113575988745212014-06-01T17:30:32.450-07:002014-06-01T17:30:32.450-07:00Step ( 9 ) isn't clear for the infinite desce...Step ( 9 ) isn't clear for the infinite descent in step (8) as it is not explained how the smaller square - P^2, has the same properties.<br />This is, I believe the missing bit :-<br />In step (6) it states ab and a^2+b^2 are both squares. In the same way, since a and b are relatively prime, it follows that a and b are both squares. So a=Q^2 and b=R^2 for integers Q, R.<br />Therefore Q^4 + R^4 = P^2<br />This is what step (9) refers to. I hope this has helped clarify the solution.<br />Jezhttps://www.blogger.com/profile/03222601835879357192noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-24120682599302270872014-05-17T23:53:46.042-07:002014-05-17T23:53:46.042-07:00how about this,,, I counted and then conclude
a^4 ...how about this,,, I counted and then conclude<br />a^4 + b^4 = C^4 - 2(ab)^2<br />what this equation same as (xn ') 4 + (y') 4 = (z (2n ')) 2?I.Bhttps://www.blogger.com/profile/02692571516267669278noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-421076580625163452014-05-08T10:31:29.119-07:002014-05-08T10:31:29.119-07:00If I wanted to prove something for, let's say ...If I wanted to prove something for, let's say a n of 8, or 16...how could I go about that argument??Dorthee Bermanhttps://www.blogger.com/profile/16974801214866358573noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-89707723283261153302013-07-29T05:54:43.217-07:002013-07-29T05:54:43.217-07:00OK, Fermat proved that the equation raised to the ...OK, Fermat proved that the equation raised to the fourth power could not generate whole solutions using his theory of limitless descent, but it was Andrew Wiles who proved the remainder of the theorem in 1986. His proof is final and superior to all others which attempt to prove the theorem. It shows that there are no valid integer solutions for x*y*z<> 0 for all equations in x^n+y^n=z^n for all values greater than 3 and truly verifies Fermat's theorem. If you wish to learn more, read his theorem yourselves.Donec Visushttps://www.blogger.com/profile/11168953957367192354noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-4396862167052749022011-01-18T08:01:44.696-08:002011-01-18T08:01:44.696-08:00Hi Larry, in reply to Graeme, my/Fermat's vers...Hi Larry, in reply to Graeme, my/Fermat's version demonstrates why k > 2 has to be 3 minimum, which is proved by implication that D = (x^3)^1 + (y^3)^1 = (z^3)^1 has no solutions. To prove FLT all we do is let k be any integer > 2 (not > 1) in B and C, and by implication D will never have any solutions. I agree that no proof of FLT occurs when k > 1 is 2 minimum, because D reduces to Pythagoras theorem. It is explained below why this "flaw" is a paradox, which Fermat resolves when working through the details of his proof. <br /> My proof is part of a much larger conjectural reconstruction of FLT. To understand some of Fermat's possible thought processes when playing around with powers of numbers, including his "Eureka" moment "...which this marvellous proposition truly explains", the following should be added in my version after "...had no solutions". <br /> "Because values of z^4 can be written as z^2, when k = 1, (x^4)^1 + (y^4)^1 = (z^4)^1 has no solutions. It follows that the multiples of n = 4 also have no solutions, i.e. when n = 8, 12, 16, 20, ...etc. because each of these power can be written as a fourth power, i.e. 2^8 = 4^4, 3^12 = 27^4. Now, the rules of powers allow them to be manipulated in different ways. If integer a raised to power n is a^n, then a^n raised to power k is (a^n)^k. This can be expressed as (a^n.k)^1, or (a^k.n)^1, or (a^k)^n. This means that each power can be written as a product of two powers n.k. Therefore the n.k multiples of n = 4 are 4 = 4.1, 8 = 4.2, 12 = 4.3, 16 = 4.4, 20 = 4.5, ...etc. <br /> From here Fermat simply played around with a few of the lower powers and found no solutions were possible. He asked himself how and why Pythagorean triples existed, and found that PT rearranged to (z^2)^1 - (y^2)^1 – (x^2)^1 always gave a zero answer. When applied to higher powers he found triples that only came close to, but not zero, i.e. when n = 3, triples (7,6,5), (9,8,6), (12,10,9); when n = 4, triples (3,2,1) and (9,8,7); and when n = 5 and higher, triples (3,2,1). Surprisingly, this proof by 'scientific experiment' demonstrated that no power greater than squares could be separated into two powers. But this was not a 'vigorous' proof demanded by the mathematical community. <br /> With this in view, Fermat went on to list all the powers n and multiples k of n = 3, 5, 6,...etc. that had no solutions. He included the multiples of n = 2 because something did not seem right: <br />n n.k multiples <br />4 4.1 4.2 4.3 4.4 4.5... <br />2 2.1? 2.2 2.3 2.4 2.5...<br />3 3.1 3.2 3.3 3.4 3.5...<br />5 5.1 5.2 5.3 5.4 5.5...<br />6 6.1 6.2 6.3 6.4 6.5...etc<br /> He was right because there is a paradox. If all powers (n = 3,4,5,6,... etc., including n = 2) do not have solutions, then neither do their multiples (n.k). But the first multiple of n = 2, (2.1) has solutions (as proved by Pythagoras theorem) but the second, third etc multiples, 2.2, 2.3, 2.4,...etc. do not. He resolved this paradox by equating multiples 2.2 to 4.1, and 2.3 to 3.2... etc. all of which had no solutions. But it was something he had to be careful of later on. <br /> His "Eureka" moment came when he noticed that n.k: 12.1 is the lowest common multiple (LCM) of n.k: 4.3 and n.k: 3.4, This meant that as the first multiple of n = 12 and the third multiple of n = 4 had no solutions, then the fourth multiple of n = 3 also had no solutions. Then by implication the first multiple of n = 3 had no solutions. From the way powers can manipulated Fermat realized that as any k > 2 can be a multiple of n = 4, then it would imply that (x^k)^1 + (y^k)^1 = (z^k)^1 had no solutions, and did not need to calculate the LCM at all. <br /> Continue: "Fermat vigorously proved his theorem..."<br />regards PhilPhilhttps://www.blogger.com/profile/09795897126057569499noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-18167618358745670712010-11-03T15:22:54.219-07:002010-11-03T15:22:54.219-07:00Problem solved... I'm no longer troubled.
Wi...Problem solved... I'm no longer troubled.<br /><br />With the help of "lebesgue", on the NRICH forum, I now understand that equation (2b) y^2=p^2-q^2 is enough to establish that q must be even. This is a consequence of y^2 being odd, and thus one more than a multiple of 4.<br /><br />Larry, since your goal is for your proof to be accessible to amateur mathematicians, maybe a little comment just after step 2 is in order to the effect that p is odd and q is even as a consequence of the 2nd equation in step 2.<br /><br />Also, you might point out at the beginning that x is chosen to be even, and y is chosen to be odd WLOG.<br /><br />Also, if you were to number your equations, then some of the text can be made shorter and clearer.<br /><br />An example of all of these changes can be found here: http://mathhelp.wikia.com/wiki/Proof_x%5E4_%2B_y%5E4_%3D_z%5E2_has_no_solutionsGraemehttps://www.blogger.com/profile/10454086142737044344noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-45304579241486740392010-11-03T11:37:37.695-07:002010-11-03T11:37:37.695-07:00After more thought, I'm still very troubled. S...After more thought, I'm still very troubled. Slapping absolute value bars around the right hand side of equation<br /><br />(2b) y^2 = p^2 - q^2<br /><br />makes the next step problematic. In order for this equation to yield a new Pythagorean triple, p must be greater than q. In order to use q=2ab later, q must be even.<br /><br />Interchanging p and q won't work, because the larger of the two must be odd.<br /><br />I'm sure I'm missing a way to cope with the smaller of p,q being odd, but I can't see it. The infinite descent escalator has ground to a halt.Graemehttps://www.blogger.com/profile/10454086142737044344noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-72401885483001674972010-11-03T10:21:50.436-07:002010-11-03T10:21:50.436-07:00Larry, I'm troubled by the selection of p,q in...Larry, I'm troubled by the selection of p,q in step 2 such that q is even and q<p. If you slapped a pair of absolute value bars around the second equation in that step, I would be less troubled.Graemehttps://www.blogger.com/profile/10454086142737044344noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-3632819647260229642010-11-03T08:22:09.484-07:002010-11-03T08:22:09.484-07:00Regarding Phil's "proof" of FLT...
I...Regarding Phil's "proof" of FLT...<br />I will recap Phil's argument here:<br />Let A=(x^4)^1+(y^4)^1=(z^4)^1<br />Let B=(x^4)^k>2+(y^4)^k>2=(z^4)^k>2<br />Let C=(x^k>2)^4+(y^k>2)^4=(z^k>2)^4<br />Let D=(x^k>2)^1+(y^k>2)^1=(z^k>2)^1<br />As A had no solutions, and A implied B, B had no solutions. <br />As B equalled C, C had no solutions. <br />As D implied C, D had no solutions. <br /><br />Rather than point out flaws, I will restate it with k>1 rather than k>2:<br /><br />Let A=(x^4)^1+(y^4)^1=(z^4)^1<br />Let B=(x^4)^k>1+(y^4)^k>1=(z^4)^k>1<br />Let C=(x^k>1)^4+(y^k>1)^4=(z^k>1)^4<br />Let D=(x^k>1)^1+(y^k>1)^1=(z^k>1)^1<br />As A had no solutions, and A implied B, B had no solutions. <br />As B equalled C, C had no solutions. <br />As D implied C, D had no solutions. <br /><br />This second version of the proof is equally valid, as I'm sure Phil will agree, since the implications are the same: A->B, B=C, and D->C. Yet, it proves something which is clearly false. Therefore, there must be a flaw in both versions of the proof.Graemehttps://www.blogger.com/profile/10454086142737044344noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-80790582047036118172010-10-13T13:28:16.969-07:002010-10-13T13:28:16.969-07:00Hi Larry,
Your correspondents want to discuss as...Hi Larry, <br /> Your correspondents want to discuss aspects of your proof of n = 4, but I would like to submit a proof for FLT which may have been Fermat's own version. <br /> After playing around with Diophantus' solution to Q8 Bk II around the early 1630's, Fermat knew that Pythagorean triples could be described algebraically in terms of a power 2 raised to a unit power by the equation (x^2)^1 + (y^2)^1 = (z^2)^1, when x,y,z are integers. From this he found the equation (x^n)^k + (y^n)^k = (z^n)^k described any power separated into two powers, when power n is raised to power k. Your first conjectural reconstruction of "The easiest proof for Fermat's Last Theorem is the case n = 4...." concerns Fermat's marginal note to Q29 Bk V of the Arithmeticorum annotated around the mid 1630's, saying that he had proved x^4 + y^4 = z^2 had no solutions. Because values of z^4 can be written as z^2, it followed that (x^4)^1 + (y^4)^1 = (z^4)^1 had no solutions. <br /> Fermat vigorously proved his theorem by applying the logic of mathematical proof introduced by the ancient Greeks: <br />Let A=(x^4)^1+(y^4)^1=(z^4)^1<br />Let B=(x^4)^k>2+(y^4)^k>2=(z^4)^k>2<br />Let C=(x^k>2)^4+(y^k>2)^4=(z^k>2)^4<br />Let D=(x^k>2)^1+(y^k>2)^1=(z^k>2)^1<br />-As A had no solutions, and A implied B, <br />B had no solutions. <br />- As B equalled C, C had no solutions. <br />- As D implied C, D had no solutions. <br />Replace (k>2) with any integer >2 in equations B, C, and D, and FLT (D) had no solutions when n>2 and k=1. <br /> In the early 1650's, Fermat derived his second proof of n = 4 by proving Pythagorean triangles do not have square areas, as shown in your reconstruction.<br /> The lesson not learned by all later mathematicians from Euler to Wiles, and even now in 2010, is that Fermat proved his theorem true for all n>2 by extending the case for n=4, not by taking different approaches to odd prime values of n. <br />Phil CutmorePhilhttps://www.blogger.com/profile/14675857755049695726noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-83789233058497731032010-06-03T01:57:28.417-07:002010-06-03T01:57:28.417-07:00Larry
Could I suggest that you expand on step (6)...Larry<br /><br />Could I suggest that you expand on step (6) a little to explain that ab and a2 + b2 are relatively prime because p and q are relatively prime?<br /><br />Could you also state that p and q are relatively prime in step (2), as you do in step (4)?<br /><br /><br />Sorry if it seems as though I want to be led fully down the path without having to think too much for myself on these proofs.<br /><br />But I <b>do</b>. ;-)<br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-32861122654922341942010-06-03T01:44:28.799-07:002010-06-03T01:44:28.799-07:00Morning Larry.
(Well it is morning here.)
In step...Morning Larry.<br />(Well it is morning here.)<br /><br />In step (7), to begin the infinite descent, don't we need an equation of the following form?<br /><br />x^4+y^4=z^2<br /><br />I finally figured out that in step (6) ab is proved to be a square.<br /><br />Now because a and b are relatively prime this means that a and b are both squares, so:<br /><br />a=G^2, b=H^2<br /><br />Which gives us in step (7):<br /><br />G^4+H^4=P^2<br /><br />Which is the same form as the original equation and the infinite descent now <i>clearly</i> works.<br /><br />I love this proof. It is so simple and elegant.<br />(Once you can get your head around the concept of infinite descent.)<br /><br />Love the blog as well. :-)<br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-10330699517167756622010-04-05T15:32:48.228-07:002010-04-05T15:32:48.228-07:00Michael,
You are correct about Wile's proof. ...Michael,<br /><br />You are correct about Wile's proof. <br /><br />In this blog, I am showing historical results that preceded Wile's master proof.<br /><br />-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-60047079512881683072010-04-05T15:31:25.068-07:002010-04-05T15:31:25.068-07:00Hi Matt,
Step #8 is the critical step which compl...Hi Matt,<br /><br />Step #8 is the critical step which completes the proof. If a positive integer solution always implies that a smaller integer solution also exists, then there can be no integer solution.<br /><br />The reason is that integers cannot descend infinitely. That's the principle behind proof by infinite descent.<br /><br />-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-44555214023463197112010-04-05T15:18:54.293-07:002010-04-05T15:18:54.293-07:00Fermat's Last Theorem is true. This was proven...<i><br />Fermat's Last Theorem is true. This was proven by Andrew Wiles.</i><br />to be more precise, he <i> finished </i> the proof by proving that semistable elliptical curves can not be modular. (It had already been shown that an integer solution to n>=3 would imply the existence of modular semistable elliptic curve.)Michael Ejercitohttps://www.blogger.com/profile/10707862691472293497noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-32863890187064245582010-04-05T15:00:55.749-07:002010-04-05T15:00:55.749-07:00Hey,
In step 8, why does it matter that all tho...Hey, <br /> <br />In step 8, why does it matter that all those numbers are smaller than each other? arent they suppossed to me? Just curiousMatthttps://www.blogger.com/profile/14096318514782286567noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-56389961714640487652010-04-05T14:46:40.340-07:002010-04-05T14:46:40.340-07:00Hi John,
This follows from step #6 where I show t...Hi John,<br /><br />This follows from step #6 where I show that a^2 + b^2 is itself a square.<br /><br />That is, there is a number P such that P^2 = a^2 + b^2.<br /><br />-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-49795851893671739432010-04-05T14:43:07.462-07:002010-04-05T14:43:07.462-07:00can you explain line 7 ? i dont understand where ...can you explain line 7 ? i dont understand where the p^2 came fromjohnhttps://www.blogger.com/profile/03583613708800665088noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-65746844060784197782009-08-15T16:18:33.529-07:002009-08-15T16:18:33.529-07:00Hi Dirk,
I appreciate the question.
From a mat...Hi Dirk,<br /><br />I appreciate the question. <br /><br />From a mathematical viewpoint, your reasoning would not be valid.<br /><br />Mathematical proofs focus on assumptions and conclusions that are directly derived.<br /><br />Saying FLT is true is really saying that given a certain set of assumptions, FLT is true.<br /><br />This is what happens for example with NonEuclidean Geometry where certain Euclidean theorems still hold up but others (such as the theorem that a triangle's angles add up to 180) do not.<br /><br />For these reasons, even if we know that sqrt(2) is not rational and FLT is true, it may turn out that there is an interesting proof that shows that if FLT is false, sqrt(2) is rational. <br /><br />My point is that the statement of two true theorems is not enough. The relationship requires its own proof in order to be valid.<br /><br />Cheers,<br /><br />-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.com