The "Last Riddle" of Pierre de Fermat, II Alexader Mitkovsky mitkovskiy@gmail.com Some time ago, I published a work etitled, "The Last Riddle" of Pierre de Fermat " i which I had writte a proof of the theorem. (See (http://wbabi.et/mitkovsky/mitkovsky6.pdf, i Eglish, ad also, http://wbabi.et/mitkovsky/mitkovsky6r.pdf for the Russia versio.) The proof was writte as mathematical formulas, but the priciple of the proof, it seems, was ot clear to the mathematicias, who studied the work. Therefore i the preset paper, I shall state the priciples ivolved i the theorem i more detail, so that it is more easily uderstood. Probably the reaso that mathematics was uable to prove Fermat's Last Theorem over a exteded period of time, is that it was assumed to be very complicated. Actually, the task is so simple that the time spet o a descriptio of the proof is much loger tha the proof of the theorem itself. To uderstad this, it is ecessary to follow Pierre de Fermat's trai of thought, ot to prove the theorem, but to deduce it. To ay ma havig a certai mathematical tur of mid ad a spatial imagiatio, it is possible to uderstad the reasoig which leads to the coclusios. For this purpose we shall temporarily overlook the fact that there is such a problem ad cocetrate o the Pythagorea Theorem. We kow that a square ca be established as the sum of two squares o a right-agled triagle. Let's accept, that the oe side equals z, ad the other sides are equal to x ad y 1, so that x +y 1 =z (1) Formula (1) meas that the area of the square z is equal to the sum of the areas of two squares, oe of which has the side x, ad the other, side y 1. We shall ot cosider all the variats of the decompositio of squares; we shall cosider oly oe of them i which z ad x are whole positive umbers. It is kow that, if z ad x are whole positive umbers, y 1 ca be whole or a irratioal umber. This was kow ceturies before B.C., ad we may assume it was also kow by Pierre de Fermat. Let's assume further that Pierre de Fermat, readig "Arithmetica", i which Diophatus of Alexadria described the tasks of decompositio of squares, had asked himself a questio - "What will happe if i the trasformatio we add oe more measuremet height?". Let's aswer this questio. Let's add to a square, the property z for height, creatig a cube with a volume of z 3. If to the squares o which z, is displayed, that is x ad y 1 we add the same height, we shall see, that the cube with volume z 3 cosists of two geometrical figures a direct parallelepiped, o which lays a square with side x ad havig height z, ad a direct parallelepiped, with side y 1 havig height z. Further we shall assume the parallelepiped has straight lies, sice all parallelepiped s, used below, have straight lies. Let's write dow the formula:
x z+y 1 z=z 3 () Formula () follows from the Pythagorea Theorem ad is ot subject to doubt. Let's cosider the compositio of formula () Here we have accepted that z ad x, are whole positive umbers ad it follows that the volume of a cube with the height z is a whole positive umber, ad the volume of the parallelepiped x z is also whole positive umber. If y 1 is a whole positive umber, the volume of the parallelepiped with the base y 1 ad height z will be atural umber. If y 1 is a irratioal umber, the volume of the parallelepiped with the base y 1 ad height z will be a irratioal umber. It is clear that a irratioal umber squared ad multiplied by a iteger, caot be a atural umber. It is obvious, that this reasoig o the decompositio of oe volumetric figure to two other volumetric figures, will be valid for ay height that is set, provided that this height is a iteger. Proceedig from this, istead of height z, we ca substitute height z -. Let's write dow the formula: x z - +y 1 z - =z (3) Where is a atural umber. Formula (3) shows that havig chaged the height of z to z -, istead of cube z 3, with z height, we have a parallelepiped with height z - ad a volume equal to z. The parallelepiped with volume z, i formula (3) may bee spread over two parallelepipeds. The base of the first parallelepiped with side x ad height z -, gives a volume equal to x z - ad this volume expressed umerically, is a atural umber. The base of the secod parallelepiped with side y 1 ad height z -, its volume is equal to y 1 z -. With z - as a atural umber, it is obvious, that if y 1 is a irratioal umber, the volume y 1 z - is also irratioal. However, for us, the parallelepiped x z - is uiterestig, as those are more importat which will accept x i trasformatios. Therefore, we ca allocate from parallelepiped x z -, aother parallelepiped, i which the square with side x has a height x -. The volume of such a parallelepiped will be equal to x. Thus, we have divided parallelepiped x z - ito two parallelepipeds, oe with the base x ad height x -, whose volume is equal to x, ad a secod parallelepiped with base x ad height z - -x - whose volume is equal to x (z - -x - ). The total of parallelepiped z will be equal to z =x + x (z - x - ) + y 1 z - (4) Formula (4) shows, that the parallelepiped with the base z, ad height z - ad volume z cosists of three parallelepipeds x, x (z - -x - ) ad y 1 z -. The above-metioed derivatios are obvious ad directly follow from the Pythagorea Theorem. We have reached that part of our reasoig i which we ca already make assumptios. For example, we ca assume that there is a expressio:
z = x + y (5) Let's cosider a situatio which ca accept y icorporatig the above coditios, i.e., whe z ad x are atural umbers. For this purpose we apply formula (4). I this formula we already have deduced the meaigs of z ad x. If we compare formulas (4) ad (5), it turs out that y should be equal to the subtractio of x from z. Let's write dow the formula: y = z - x = x (z - x - ) + y 1 z - (6) Formula (6) shows, that the meaig y should cosist of the sum of the volumes of two parallelepiped, x (z - -x - ) ad y 1 z -. Earlier we have established that the volume of parallelepiped x (z - -x - ) is always a atural umber, ad the volume of parallelepiped y 1 z - ca be a atural umber if y 1 is a atural umber, or irratioal if y 1 is a irratioal umber. It is easy to uderstad that if oe is a irratioal umber, ad the secod is a atural umber, the sum of these will be a irratioal umber. This implies, that y ca be a atural umber oly i the evet that y 1 is a atural umber. It is obvious, that x, y 1 ad z simultaeously will be itegers oly i the evet that x, y 1 ad z are a Pythagorea triple. Let's cosider, i what cases this coditio ca be carried out. 1. Lets assume that z is a eve umber. We kow that it is impossible to spread out a square whose property is a eve atural umber o two whole squares. This implies that at ay eve meaig z, the meaig y 1 will always be a irratioal umber, provided that z ad x are atural umbers. Hece, the volume parallelepiped y 1 z - ca ot be expressed by a atural umber, meaig from formula (6), y ca ot be a atural umber. It is thus obvious, that y ca ot be atural umber.. Lets assume, that z is a simple odd umber. It is kow, that if the simple umber looks like 4t+1, this will be a Pythagorea umber. It is obvious, that i this case, y 1 will be atural, ad the volume of the parallelepiped y 1 z - will be expressed as a atural umber. This implies that the meaig y (formula 6) will be a atural umber ad y ca be a atural umber. Let's accept, that y 1 is a atural umber. The meaig of y i this case will be a atural umber, which follows from formula (6). This meas y ca preset a parallelepiped, i which the base is the square with side y, havig height y -. Square y ca be etered i the base z of parallelepiped z. I this case, from the Pythagorea theorem, there should be a meaig for x 1, which will satisfy the coditio x 1 +y =z (7) Usig formula (7), the parallelepiped z ca be preseted as the sum of two parallelepipeds, oe with base x 1 ad height z -, ad the secod with base y ad height z -. x 1 z - +y z - =z (8) From the parallelepiped y z - we shall derive a parallelepiped with base y ad height y -. We shall determie that parallelepiped y z - cosists of two parallelepipeds, oe with base y, height y -, volume y, ad oe with base y, height z - -y -, ad volume y (z - -y - ). I total, parallelepiped z will make: z =y + y (z - y - ) + x 1 z - (9)
From formulas (5) ad (9), it follows, x = z - y = y (z - y - ) + x 1 z - (10) From formula (10) follows that if x 1 is a irratioal umber, x ca ot be a atural umber. That cotradicts oe of the coditios. The ecessary coditios follow from above, that x ad y will be atural: That side y is atural, the it is ecessary that y 1 is a atural umber. That side x is atural, the it is ecessary, that x 1 is a atural umber. These also are ecessary coditios. So it is satisfied as a coditio of itegrity that for x ad y for oe z there should be two Pythagorea triples, based o z. We kow, that simple odd umbers of the kid, 4t+1 are represeted as the sum of squares i a uique way. This is easy to prove. Ay oe primitive Pythagorea triple (x, y, z) is uequivocally represeted as m - ; m; m + Where z=m +, m ad are atural ad mutually simple umbers. We have accepted, that z is a simple odd atural umber. From the theorem of Fermat Euler, it is kow that the simple odd umbers of the kid 4t+1 are represeted as the sum of squares i a uique way, hece, for the equality z=m + there is oly oe pair of umbers, m ad. This implies, that we ca have oly oe meaig of x=m - ad y =m. Hece, we have o other variats i the performace of ecessary coditios of atural umbers, except as i the case x=x 1 y=y 1. Did Pierre de Fermat kow about this depedece? Udoubtedly! It is kow that it is possible to preset the statemet that a simple umber is the sum of two whole squares oly i the evet that this umber is odd. More precisely, it shows that divisio by 4 gives 1 i the remaider. This had bee formulated by Fermat i 1640 ad the, after a hudred years, this statemet was prove by Euler. It is kow to us as the Fermat Euler theorem. Thus, we came to the coclusio that x ad y i equatio (5) were atural umbers, ad the coditio x=x 1 y=y 1 is ecessary. I this case equatio (4) will accept z =x + x (z - x - ) + y z - y < y z - This implies x + y x + x (z - x - ) + y z - I this case, equatio (4) will accept z =y + y (z - y - ) + x z -
x < x z - This implies x + y y + y (z - y - ) + x z - Thus, we came to the coclusio that ay simple odd atural umber z of the kid 4t+1 does ot exist i the atural values x ad y, i which there is a resolutio of the equatio z = x + y. It is obvious that if a odd z does ot look like 4t+1, this umber caot be spread out to a Pythagorea triple. 3. We shall assume, that z is a compoud umber, ad desigate it as z 1 = zk, where z is a simple odd atural umber, k is a atural umber. Let's icrease this umber to degree, to get the umber z 1, which ca be preseted as a parallelepiped i which base, the square has the lie z 1 ad the height z 1 -. We admit, we eed to spread z 1 over two umbers, x ad y uder the formula z 1 = x + y (5) Let's accept that x is a atural umber, which ca be preseted vectorally as a parallelepiped with the base x ad height x -. Parallelepiped x is a part of parallelepiped z 1, ad the base x is a compoet of the base z 1. However, parallelepiped z 1 is also possible to be preseted as z 1 =( zk) = z k = z z - k (11) Formula (11) shows that umbers of the kid z 1 ca be preseted vectorially as a parallelepiped, with base z ad height z - k, with the side of the square havig z as a simple odd atural umber. Obviously, as i this case, x will be a part of parallelepiped z 1, but the base of parallelepiped x will be aother ad this base will be a part of the base z. As we have accepted, that - x is a atural umber, we ca accept the base of parallelepiped x as a square, whose side is atural whole positive umber, which we shall desigate x 1. I this case, the volume of parallelepiped x is possible to be preseted as x 1 a, where a is the height of parallelepiped x = x 1 a. It is ecessary to ote, that a ca be both a whole ad fractioal umber. However i ay case, a is ratioal umber. The meaig of x 1 we ca set arbitrarily, guided oly by the fact that this umber should be whole ad less tha z. From the Pythagorea Theorem there should be parity: x 1 +y 1 =z (1) we ca spread parallelepiped z 1 over two parallelepipeds - oe parallelepiped, i which the base is the square of side x 1 ad havig a height of z - k, ad a secod parallelepiped, i which the base is a square with the side y 1 havig the height z - k, provided that x 1 +y 1 =z. Thus, the volume parallelepiped z 1 will cosist of volumes of two parallelepiped so that it is possible to write dow the formula: z 1 = z z - k = x 1 z - k +y 1 z - k (13)
It is usettlig to otice, that the right ad left part of the formula are easily deduced from k, ad as a result, we have the equatio. z = x 1 z - +y 1 z - (14) I which z is a simple odd atural umber. Or, z 1 / k = z = x /k +y /k (15) It is obvious that a reductio of formula (13) is proportioal to k. The volume of parallelepiped x / k = x 1 a/ k was also reduced i such a maer that the height would decrease i k ad would be equal to height b, where b = a/ k. Thus for b, there ca be both a whole, ad fractioal umber, but i ay case, b is a ratioal umber. Parallelepiped x 1 z - ca spread over two parallelepipeds - x 1 b ad x 1 (z - -b) Where x 1 b = x / k. The volume parallelepiped z will be equal to the sum of volumes of three parallelepiped x 1 b, x 1 (z - -b) ad y 1 z - z = x 1 b +x 1 (z - -b) +y 1 z - (16) Recogizig that x 1 is a atural umber, b a ratioal umber, it follows, that x 1 b is a ratioal umber. It is possible to coclude from this, that x 1 (z - -b) are ratioal umbers. If x 1 b = x / k, from formula (15), it follows, that y /k = x 1 (z - -b) +y 1 z - (17) From formula (17) it is obvious that if y 1 is a atural umber, y /k ca be a ratioal umber. If y 1 is a irratioal umber, y /k will be a irratioal umber. from what follows, if y 1 is irratioal, the umber y ca ot be a ratioal umber. Coditios, i which y 1 is a atural umber for the odd whole positive umber z of the kid 4t+1 that we have cosidered above, we have established that for ay simple odd atural z of the kid 4t+1 there does ot exist a atural x ad y, for which there is a solutio of the equatio z = x + y. We have cosidered all possible variats. See diagram below:
The above evaluatio may be easier to uderstad pictorially, usig the followig figure: x - x z y y - z - Sice z ca ot accept other atural meaigs, we came to the coclusio that the equatio z = x + y has o solutios for whole positive umbers of >. As we see from the reasoig, the meaig of does ot ifluece the proof of the theorem, as idicated by Pierre de Fermat. Ad this really surprisig proof, is completely based o the kowledge that was available to Pierre de Fermat. I hope after due study, there will be o doubts that Pierre de Fermat kew the proof of the theorem, which was subsequetly amed the Last Theorem of Fermat. Novosibirsk Russia Alexader Mit kovskiy mitkovskiy@gmail.com