tag:blogger.com,1999:blog-12535639.post8494099071945680663..comments2021-04-02T00:42:48.223-07:00Comments on Fermat's Last Theorem: Galois' Memoir: Lemma 2 (Galois Resolvent)Larry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-12535639.post-83891541582700733112015-06-09T14:35:22.145-07:002015-06-09T14:35:22.145-07:00Hi guys, After much deliberation I have at last se...Hi guys,<br />After much deliberation I have at last seen the simplest and shortest proof of FLT which is so simple that it is laughable that it has taken us all so long to find it! <br /> Assume Z^n = Y^n + X^n exists.<br /> Multiply through by Z^n to give (Z^n)^2 = (Z.Y)^n + (Z.X)^n = (Z^2n)^1.<br /> Multiply through by Z^2n to give (Z^2n)^2 = (Z^3.Y)^n + (Z^3.X)^n.<br /> Multiply through by Z^2n to give (Z^2n)^3 = (Z^5.Y)^n + (Z^5.X)^n.<br /> We can continue like this ad infinitum to give every (Z^2n)^m. Every (Z^2n)^m forms a Pythagorean triple, not necessarily primitive so from step  onwards we have a one to one correspondence between the Fermat equation and the Pythagorean equation. So we can only conclude that the only possible value for n is 2 so that our assumption in  is an absurdity. Q.E.D. <br />nb: The simple procedure of splitting the square generates the Pythagorean triples from every (Z^2n)^m.<br />Although  seems very trivial it is a consequence of Euclids &#39;Elements&#39; book 2 propositions 4 &amp; 7 and which therefore constitute a proof for the &#39;Distributive Law of Multiplication&#39;.<br />It is also a consequence of the &#39;Unique Composite Equation for the Triple of the Powers&#39; i.e. X^n.Y^n.Z^n. which is equal to three composite equations.<br />All three methods generate three unique linked quadruples which are satisfied by the Pythagorean triples as inputs and since nobody else seems aware of them I have called them Fermat-Bateman quadruples.<br />81^2 = 9^3 + 18^3 at first sight seems to contradict  but of course 81^2 does not form a cubed integer and if it did it would prove Fermat&#39;s Last Theorem false.<br />Comments as to any flaws in my logic would be appreciated.Alastair Batemanhttps://www.blogger.com/profile/14395447443056931757noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-4364463380676835542015-04-17T20:10:03.278-07:002015-04-17T20:10:03.278-07:00Hi Guys I have found The Relations of Barlow when...Hi Guys<br /><br />I have found The Relations of Barlow when I read Ribenboim&#39;s Book. &quot;Playing&quot; with it, I found another sistem of equations, and then, I used modular congruences and I found what is summarized below (we can work just with natural numbers to analize the fermat&#39;s integer soluctions,because if one or 2 of the numbers are negatives, them we can change the place in the equation in order to have just positive numbers. If al of them are negative, we can multiply the equation by -1):<br /><br />I didn&#39;t make an English version*, because I am having dificulties in the mahts expression (loses its format when I paste here), but if you follow the steps below, then you will achiev the same results :<br /><br />* I published the entire demonstration in a Blog I created (in this blog is better to read the maths equations) and in other forums in Brazil and others countries :<br /><br />http://filosofbeer.blogspot.com.br/2015/04/o-ultimo-teorema-de-fermat-as-relacoes.html<br /><br />From The Barlow Relations (See Fermat Last Theorem for Amateurs, pg 99 - but is very easy to show this) we have, in both Case 1 (n dont divide z) and Case 2 (n divide z) of FLT :<br /><br />z-x = y0^n<br />z-y = x0^n<br /><br />From this , we conclude thar<br /><br />y-x = y0^n-x0^n = (y0-x0)(y0^(n-1)+....+x0^(n-1))<br /><br />gdc(y-x, y0^(n-1)+....+x0^(n-1)) = y0^(n-1)+....+x0^(n-1)=p<br /><br />y0^(n-1)+....+x0^(n-1) &quot;=&quot; 0 mod (p)<br />y-x &quot;=&quot; 0 mod (p)<br /><br />Replacing x0 = (z-y)^(1/n) and y0 = (z-x)^(1/n)<br /><br />[(z-y)^(n-1)]^(1/n)+....+[(z-x)^(n-1)](1/n) &quot;=&quot; 0 mod(p)<br /><br />Replacing y &quot;=&quot; x mod (p)<br /><br />n[(z-x)^(n-1)](1/n) &quot;=&quot; 0 mod(p)<br /><br />ny0^(n-1)&quot;=&quot; 0 mod(p)<br /><br />p = 1 or p divides y0^(n-1) or p=n<br /><br />As x and y are both &gt; 0, then y0^(n-1)+....+x0^(n-1) is always &gt; 1. So p can not be 1<br /><br />If p divides y0, then it will divide x. But gdc(x,y,z)=1, so p can not divide y0 or x0<br /><br />If p=n, then<br /><br />y0^(n-1)+....+x0^(n-1)= n<br /><br />But this sum has n terms, and this is possible just if y0=1 and x0 = 1 , because both are &gt; 0. And if one of them is larger then 1, then the sum wil be larger than n.<br /><br />If x0=1 and y0=1, then using the Relations of Barlow, we conclude that x=y, wich contradicts te conditions z&gt;y&gt;x and gdc (x,y)=1.<br /><br />The same analysis could be used in the case 2 (even if n divedes x ou y).<br /><br />So, there&#39;s no integers soluction for the equation x^n+y^n = z^n<br /><br />FelipeRJ<br /> <br />Anonymoushttps://www.blogger.com/profile/09443873579985260509noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-74880984208824939972014-03-21T14:42:35.562-07:002014-03-21T14:42:35.562-07:00To Thao Tran. &#39;Is it interesting?&#39; Well fo...To Thao Tran. &#39;Is it interesting?&#39; Well for me personally it&#39;s a little, no a lot over my head as I am limited to OLD high school math&#39;s so I honestly couldn&#39;t say. Sorry.Alastair Batemanhttps://www.blogger.com/profile/14395447443056931757noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-15792048068002680462014-03-03T01:01:47.031-08:002014-03-03T01:01:47.031-08:00There is another explanation of a simple proof of ...There is another explanation of a simple proof of Fermat’s last theorem as follows:<br />X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime &gt;2) (1)<br /> <br />1. Let‘s divide (1) by (Z-X)^p, we shall get:<br /> (X/(Z-X))^p +( Y/(Z-X))^p ?= (Z/(Z-X))^p (2)<br /><br />2. That means we shall have:<br /> X’^p + Y’^p ?= Z’^p and Z’ = X’+1 , with X’ =(X/(Z-X)), Y’ =(Y/(Z-X)), Z’ =(Z/(Z-X)) (3)<br /><br />3. From (3), we shall have these equivalent forms (4) and (5):<br />Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)<br />Y’^p ?= p(-Z’) ^(p-1) + …+p(-Z’) +1 (5)<br /><br />4. Similarly, let’s divide (1) by (Z-Y) ^p , we shall get:<br />(X/(Z-Y)) ^p +( Y/(Z-Y)) ^p ?= (Z/(Z-Y)) ^p (6)<br /><br />That means we shall have these equivalent forms (7), (8) and (9):<br />X” ^p + Y” ^p ?= Z” ^p and Z” = Y”+1 , with X” =(X/(Z-Y)), Y” =(Y/(Z-Y)), Z” =(Z/(Z-Y)) (7)<br /><br />From (7), we shall have:<br />X” ^p ?= pY”^(p-1) + …+pY” +1 (8)<br />X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)<br /><br />Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).<br /><br />By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X” ^p + Y” ^p ?= Z” ^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X’’ ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.<br />X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:<br /><br />i) In (8) and (9), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1), that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY” ^(p-1)) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.<br />Fermat’s last theorem is simply proved!<br /><br /> <br />ii. With X^p + Y^p ?= Z^p , if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:<br />We should have : X^p + Y” ^p ?= Z” ^p , then X” ^p ?= 2Z” ^p or (X”/Z”) ^p ?= 2. The equal sign, in (X”/Z”) ^p ?= 2, is impossible.<br />Fermat’s last theorem is simply again proved, with the connection to the concept of (X”/Z”) ^p ?= 2. Is it interesting? <br />Email: thaotrangtvt3@gmail.com<br /><br />Anonymoushttps://www.blogger.com/profile/12613121951684904144noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-19214905967219131762014-02-01T12:46:24.026-08:002014-02-01T12:46:24.026-08:00I am re-instating my original proof as valid with ...I am re-instating my original proof as valid with the following sentence inserted after REDUCTIO AD ABSURDUM. &#39; But a^n•a^n - b^n•b^n = (a^n + b^n)•a^n - (a^n + b^n)•b^n is valid since: (a^n•a^n + a^n•b^n) - (a^n•b^n + b^n•b^n) = a^n•a^n - b^n•b^n since the two &#39;strike through&#39; terms cancel each other out. It has therefore been demonstrated that (a^n + b^n) only has an integer solution and can never equal C^n. &#39;.Alastair Batemanhttps://www.blogger.com/profile/14395447443056931757noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-21821180855244717762013-11-26T09:08:30.955-08:002013-11-26T09:08:30.955-08:00I forgot I had posted the comment. It is of course...I forgot I had posted the comment. It is of course wrong because the equation (a^n)^2 - (b^n)^2 = C^n(a^n -b^n) is VALID!! C^n can seem to be equal to a^n &amp; b^n at the same time because there is a profundity here that is not apparent at first sight. Since C^n = a^n + b^n then a^n(a^n + b^n) - b^n(a^n+b^n) = a^n + b^n ( what I call the &#39;escape knot equation&#39; ) so we are back where we started and can go round the loop endlessly.( &#39;Infinite Descent&#39;?) What this demonstrates is that all x^n ultimately collapse to a solution in the squares and as far as I can see it was the observation that the generic equation x^n = z^n2- y^2 is obtained when x^n is split because the algorithm is precisely the one used to split a square into two squares which is where Fermat made his observation in &#39;Arithmetica&#39;. The result is always an all integer solution for x,y,z and the exponents but only two of the exponents are the same power which is always 2 so Fermat&#39;s statement was true, so yes he did have a proof contrary to what present day pundits tell us. That means as far as I am concerned that the quest for the modern proof is founded on the lie that Fermat did not have one. I have put a revised proof as a video on YouTube called &#39;Simpleton Proof of FLT&#39; for the benefit of those who like me only understand high school maths. Alastair Batemanhttps://www.blogger.com/profile/14395447443056931757noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-73947591592910883442013-08-09T13:22:38.675-07:002013-08-09T13:22:38.675-07:00The Marginal Proof of Fermat&#39;s Last Theorem S...The Marginal Proof of Fermat&#39;s Last Theorem<br /><br />Step[i] Assume C^n = a^n + b^n exists for 4 natural / integer numbers C, a, b &amp; n.<br />Step[ii] The difference of two squares is given by the well proven identity <br />x^2 - y^2 = (x + y)•(x - y) : Euclids Elements : Book 2 : proposition 5 &amp; 6.<br />Step [iii] x &amp; y can be any of the infinity of the natural numbers so let x = a^n and y = b^n therefore (a^n)^2 - (b^n)^2 = (a^n + b^n)•( a^n - b^n) hence a^n•a^n - b^n•b^n = C^n•( a^n - b^n).<br />By the Distributive Law of Multiplication: a^n•a^n - b^n•b^n = C^n•a^n - C^n•b^n <br />But C^n cannot be equal to both a^n &amp; b^n at the same time <br />hence REDUCTIO AD ABSURDUM.<br />Hence the equation C^n = a^n + b^n has no reality when all four variables are natural / integer numbers as Pierre De Fermat stated all those years ago. Q.E.D. <br />As Fermat said &quot;remarkable&quot; because it is simple, absolute in what it proves, would almost fit in the margin of and is in perfect mathematical accord with where it was found in &#39;Arithmetica&#39;.<br />A.R.Bateman : arb2013@talktalk.netAlastair Batemanhttps://www.blogger.com/profile/14395447443056931757noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-72688919901622944892011-11-06T06:34:57.588-08:002011-11-06T06:34:57.588-08:008. P is symmetric w.r.t.variables A,B,C, . . . (t...8. P is symmetric w.r.t.variables A,B,C, . . . (the roots A&#39;, B&#39;, . . ) or w.r.t. a,b,c,. . . (the &quot;roots&quot; of f(x)?<br /><br />12. Why is each factor of P a nonzero polynomial?pdaboushttps://www.blogger.com/profile/15870823075637167772noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-30714537370356482182011-06-03T03:08:04.922-07:002011-06-03T03:08:04.922-07:00I Sugesh krishna C.P. Thiruvanathapuram,Kerala ,IN...I Sugesh krishna C.P. Thiruvanathapuram,Kerala ,INDIA have found a simple proof for Fermat&#39;s theorem. My e-mail id is sugeshkcp@gmail.comSugeshhttps://www.blogger.com/profile/03447656042909873128noreply@blogger.com