Jiang, Run

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The part of Fermat's theoremAuthor: by Run jiang email:jiangrunyun@outlook.comMailing address:(In china)四川广安武胜飞龙高中蒋先伦收-638400College graduate I major in Microelectronics science and technologyAbstract:We can't have an equation where the left hand side is 4 (2C+1)and the right hand side is 8C.The more detailed properties of odd and evennumbers will be covered later in this article.What's important is that when anumber or an equation maintains certain properties about odd and evennumbers. It will be difficult and there may be no other way to solve it.Keywords:Basic math.odd number and even number.Untenable equationMake M*2^n=(n) M is odd.So let's call that O(2n+1)2=4n(n+1)+1=8C+1 Any odd number to the second power is 8C+1Extension ak=4kC+1 when a is odd and k is even.In fact,ak=4*2dC+1 k=(d).a2+b2-2=16C when a+b=8C a,b are odd.Let a=8C+1 b=8C-1 or a=8C+3 b=8C-3 Equation was set upThere must be an even number of terms in the equation, and the O of theseterms is the lowest and the same.So,We omit the higher term of O. Just verify that it's big enough at the end.If 8k and 8k2 appear, 8k2 can be ignored when k is even.-Xm+Ym=AmWhen X plus Y has no divisor with mWe can get the following propertiesXm+Ym=(X+Y)Zm （X,Y,Z,m is odd）X+Y=(>=3) （by m>=3）Zm=Xm-1-Xm-2Y+…………-XYm-2+Ym-1Zm=Xm-1-Xm-2Y+…………-XYm-2+Ym-1=(X+Y)C+mYm-1Since m is odd, m minus 1 is evenYm-1-1=(>=3)When X+Y=(>=3)Zm-m=(>=3)Known m is oddEasy to knowm=4k+1. When Z=4n+1m=4k+3. When Z=4n+3（note: when m=4k+1 z=4k+3 , Zm-m=(1).m=4k+3 z=4k+1 are in sameway）Known2Zm-Xm-1-Ym-1=(X-Y)2(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3) …………（1）2(Zm-1)+1-Xm-1+1-Ym-1=(X-Y)2(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)X+Y=(>=3)Know X-Y=(1) （note X,Y are odd）By Z=4n+1 m is oddThe left-hand side is (> =3), and the right-hand side is (2).InvalidWhen Z=4n+3, m=4k+3After that, it's easy to find out about the nature of the n, k, Z, m,X.Y.Zm+1=4(n+1)(1-Z+Z2-……+Zm-1)=4(n+1)(4k+3-(Z+1)+(Z2-1)-(Z3+1)+……+(Zm-1-1))=4(n+1)((1+Z)C+4k+3)=4(n+1)(4C-1)We want to determine the fundamental properties of X,Y,n, and k（Zm=Xm-1-Xm-2Y+…………-XYm-2+Ym-1=(X+Y)C+mYm-1……+4n(4k+3)34k+2+34k+3-4k-3=(X+Y)C+m(Ym-1-1)Let's assume the right-hand side is big enough to OIn this solution,even if the value of the current X and Y combination is smallafter the O operation, we can assume that it is large. In this way, if it iswrong, it is regarded as an error, and if it is correct, it proves that the Ooperation value of X and Y is very large.……+4n34k+3+34k+3-4k-3=(>=4)So n+k+2=(>=2)So n=(1) k=(>=2) or n=(>=2) k=(1)）The following proves that n=(>=1)（We can be sure that Zm=…… is now fully established.So,we use Z2m.This is a general approach.Because there are many proofs here, for unnecessary trouble, it isnecessary to reduce the unknown.You must ensure that there is a correspondence between Zm and Xm-1, Ym-1such as,Zm-Ym-1,Zm-1+Ym-1-1+2.Because we only know“2Zm-Xm-1-Ym-1=(X-Y)2(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)”So (2Zm-Xm-1-Ym-1)(2Zm+Xm-1+Ym-1) is best choice.）(2Zm-Xm-1-Ym-1)(2Zm+Xm-1+Ym-1)=4Z2m-(Xm-1+Ym-1-2)2+4(Xm-1+Ym-1-2)-44(2C+1)(-8(n+1)+Xm-1+Ym-1-2+32C)=4*38k+6+4*4n*(8k+6)*38k+5+4*(4k+3)(8k+5)*16n2*38k+4+……-4+4(Xm-1+Ym-1-2)+128C4(2C+1)(-8(n+1)+32C+Xm-1+Ym-1-2)=4*36-32n+64(4k+3)(8k+5)n2*38k+4+……-4+4(Xm-1+Ym-1-2)+128C4(2C+1)(-8(n+1)+32C+Xm-1+Ym-1-2)=-32-32n+4(Xm-1+Ym-1-2)+128C+64(4k+3)(8k+5)n2*38k+4By X+Y=(>=3)Know (Xm-1+Ym-1-2)=(>=4)So n=(>=1) （2）（Before that, I did n’t know what the result would be. I did n’t know my goal.I could only get it closer to my goals.Of course, there must be(4n+3)8k+6=……+4n(8k+6)38k+5+38k+6 in the process.In response to use Z2m）N is determined, and there are no other properties（There are also many proofs below.We deliberately make 4-2Xm-1-2Ym-1appear in the equation.Found that it is beneficial to increase the value ofthe O operation result of 4-2Xm-1-2Ym-1.Because one of our purposes is toincrease the value of the O operation of a known item.whena+b=(5),because a = (1) b = (1) is true, when the O operation value of bincreases, that is, b = (2) is not true and we are happy to see it.）(2Zm+Xm-1+Ym-1-4)(2Zm+Xm-1+Ym-1)=4Z2m-4+2(Zm+1)(Xm-1+Ym-1-2)+(Xm-1+Ym-1-2)2+4-2Xm-1-2Ym-1（by when a+b=8C a2+b2-2=16C）32(2C+1)=4(38k+6-1)+4-2Xm-1-2Ym-1+64C32(2C+1)=4(36-1)+4-2Xm-1-2Ym-1+64C32(2C+1)=4(32-1)+4-2Xm-1-2Ym-1+64CXm-1+Ym-1-2=(>=5) （3）（I did not expect such a result.Regarding the result, we can only choose anew equation. Because the original equation can no longer produce newresults. And the new equation does not necessarily have a considerableresult. Just like before, use Z2m is to add a new equation.I have tried manytimes to get these valuable equations.）Xm-1+Ym-1-2 is determined, and there are no other properties4XY(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)+4Zm-2(Zm+1)=4(Xm-1+Xm-3Y2+……+Ym-1)-2(Zm+1)By  and Xm-1=8C+1 Xm-3Y2=8C+1 ……=32C+8k+8n+164XY(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)+4Zm-2(Zm+1)=4XY(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)+Xm-1+Ym-1-2+(2Zm-Xm-1-Ym-1)=4XY(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)+Xm-1+Ym-1-2+(X-Y)2(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)=(X+Y)2(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)+Xm-1+Ym-1-2=(>=5)In addition4XY(Xm-3+Xm-5Y2+…………+X2Ym-5+Ym-3)+4Zm-2(Zm+1)=4(Xm-1+Xm-3Y2+……+Ym-1)-2(Zm+1)=32C+8k+8n+1632C+8k+8n+16=(>=5)k+n=(1)By （2）n=(>=1)k=(>=1)k is determined, and there are no other properties（This is exactly what we imagined before.Zm=Xm-1-Xm-2Y+…………-XYm-2+Ym-1 is used here.Or,n and k appears on the left side of theequation.The right side of the equation uses the known equation to find theresult of its O operation.）Know n=(>=2) k=(1) or n=(>=1) k=(>=2)（Now, start to find the attributes of X and Y.Because it is needed later. As Isaid before, suppose the value of the O operation of the X and Y items islarge enough. Now, start to prove that the value of the operation O of X andY is large enough.）By （3） Xm-1+Ym-1-2=(>=5)When X,Y are 16C-3,16C+3 or 16C+5,16C-5 or 16C+1, 16C+7 or 16C-116C-7 ,X,Y do not meet Xm-1+Ym-1-2=(>=5)Such as X=16C-3 ,X2k+1=16C-32k+1=16C-3+3(1-32k) by k=(>=1),X2k+1=16C-3Xm-1+Ym-1-2=X4k+2+Y4k+2-2=(16C-3)2+(16C+3)2-2=32C+16So X,Y are 16C-1,16C+1 or 16C+7,16C-7 or 16C+3, 16C+5 or 16C-316C-5Know XY=16C-1Zm+1=4(n+1)(1-Z+Z2+……+Zm-1)=4(n+1)(1+Z2+……+Zm-1-Z-Z3-……-Zm-2)Zm+1=4(n+1)(8C+2k+2-Z(2k+1))=4(n+1)(8C+2k+2-(4n+3)(2k+1))By n=(>=1) k=(>=1)Zm+1=32C-4(n+1)Zm is determined, and there are no other propertiesZm+4n+5=64C+4n*34k+3+34k+3+4n+5=64C+16n(2C+1)+33(34k-1)+32Science n=(1) k=(>=2) 16n(2C+1)=32+64C (34k-1)=64CScience n=(>=2) k=(1) 16n(2C+1)=64C (34k-1)=64C+32So Zm+4n+5=64CZm=64C-4n-5（This is also necessary because it can reduce the amount of subsequentcalculations. In other words, Z=-4n-5）2(X+Y)2(Xm-3+Xm-5Y2+…………+Ym-3)(-Xm-1-Ym-1)=-(4Zm+8XY(Xm-3+Xm-5Y2+…………+Ym-3))(Ym-1-Xm-1)-8Xm-1(X(X+Y)(Xm-3+Xm-5Y2+…………+Ym-3))+(Ym-1-Xm-1)2+2(Y+X)2(Y-X)2(Xm-3+Xm-5Y2+…………+Ym-3)2-(Zm+2XY(Xm-3+Xm-5Y2+…………+Ym-3))(Ym-1-Xm-1)-2Xm-1(X(X+Y)(Xm-3+Xm-5Y2+…………+Ym-3))=(D) （4）D is a big number（Ym-1-Xm-1=(Y-X)CHere, X-Y = 16C-2 is used cleverly when X = 16C-1 or X = 16C-7 .……This is another new equation.Because at any time Zm=64C-4n-5 will causethe equation to hold. However, m can be not given directly. X, Y can appeardirectly in any equation. However, X and Y are 16C-1, 16C+1…….C is anarbitrary number, so that the O operation value of a certain two items isvery small.So, we have to consider -1 or -7 …… in 16C-1, 16C+1…… inX and Y instead of 16C）((64C+4n+5-2(16C-1)(8C+2k+1))(Y-X)((X2k+X2k-2Y2+…………+Y2k)2-X2Y2(X2k-2+X2k-4Y2+…………+Y2k-2)2)-2(8C+1)(X(8C+2k+1)))(X+Y)=(D)((64C+4n+5-2(16C-1)(8C+2k+1))(Y-X)((8C+k+1)2-(8C+1)(8C+k)2)-2(8C+1)(X(8C+2k+1)))(X+Y)=(D)((4n+5+2(2k+1))(Y-X)((k+1)2-k2)-2X(2k+1))=(>=4)((4n+5+2(2k+1))(Y-X)-2X)=(>=4) （5）X is 16C+5 or 16C-3By Y-X=16C-2 and （5）(4n+5+2(2k+1)+5)=(>=4)(4n+5+2(2k+1)-3)=(>=4)InvalidX is 16C+7 and Y is 16C-7Y-X=16C+2 X2=(16C+7)2=16C+49=16C+1By （4）-(Zm+2XY(Xm-3+Xm-5Y2+…………+Ym-3))(Ym-1-Xm-1)-2Xm-1(X(X+Y)(Xm-3+Xm-5Y2+…………+Ym-3))=(D)((64C+4n+5-2(16C-1)(8C+2k+1))(Y-X)((X2k+X2k-2Y2+…………+Y2k)2-X2Y2(X2k-2+X2k-4Y2+…………+Y2k-2)2)-2(16C+1)(X(16C+2k+1)))(X+Y)=(D)((4n+5+2(2k+1))(Y-X)-2X)=(>=5)(4n+5+2(2k+1)-7)=(>=5)InvalidSo X is 16C+1 or 16C-1Know X+Y=(>=6) （by 4k+1 is error so m=5 is error so X+Y=(5) is error）X=16C1+1 Y=16C2-1 easy to konw C1+C2=(>=1) so C1-C2=(>=1)XY=32C-1Zm=Xm-1-Xm-2Y+…………-XYm-2+Ym-1Zm=(X+Y)C+mXm-1X is 16C+1 or 16C-1Easy to know 4k+4n+8=32C Zm=64C-4n-5(4n+3)(-1+Zm-1+1)=Zm(4n+3)(1+(16n2+24n+8)(2k+1+(16n2+24n+8)((2k+1)k+8C)))=ZmWhen k=(1)32+112n+48k+48n2+144nk+3*32n2+3*64k=Zm+4n+5+512C(4n+3)(1+(16n2+24n+8)(2k+1+(16n2+24n+8)((Z2)2k-1+2(Z2)2k-2+3(Z2)2k-3+……+2k)))=Zm(4n+3)(1+(16n2+24n+8)(2k+1+(16n2+24n+8)((16n2+24n+9)2k-1+2(16n2+24n+9)2k-2+3(16n2+24n+9)2k-3+……+2k)))=Zm(4n+3)(1+(16n2+24n+8)(2k+1+(16n2+24n+8)((9)2k-1+2(9)2k-2+3(9)2k-3+……+2k+16C)))=Zm(4n+3)(1+(16n2+24n+8)(2k+1+(16n2+24n+8)((9)+2+3(9)+4+5(9)……+2k+16C)))=Zm(4n+3)(1+(16n2+24n+8)(2k+1+(16n2+24n+8)((9)k2+k(k+1)+16C)))=Zm32+112n+48k+48n2+144nk+3*32n2+64n3+3*64k(1+10k)=Zm+4n+5+1024CZm=Xm-1+Ym-1-Xm-2Y-XYm-2+…………-X2k+1Y2k+1Zm=(X+Y)2C-(4k+3)X2k+1Y2k+1XY=32C-1（This shows that Zm+(4k+3)X2k+1Y2k+1 cannot be very large after the Ooperation.Because Zm=(4n+3)4k+3 k=(>=1)……）（When a,b,m is odd.am+bm=(a+b)(am-1+……+bm-1)=(a+b)(mam-1+(a+b)C)=(a+b)(m+4(m-1)C+(a+b)C)Because am-1=4(m-1)C+1So am+bm=(a+b)m+d The O operation of d is large, when the O operationof a+b is large）（1-34k=1-(4-1)4k=16k-16*4k*(4k-1)/2+……=16k+32k+……In future calculations, please keep more terms when expanding.The next strategy is to use am+(4k+3)m(XY)m=C to get a+(4k+3)(XY). Use(4k+3)(XY)= (4k+3)(XY)2k+1+C and then change (4k+3)(XY)2k+1 to(4n+3)4k+3.In this process, there will be many single k, n, nk.First find k=(>=2), then find that k has a large in O operation, and thensubstitute(4n+3)8k+6-(4k+3)2(XY)4k+2=211C）（32nk=(>=8) so 32nk=256C）When k=(1) n=(>=2) （this can reduce trouble）32k(4n+3)4k+3+32k(4k+3)(XY)2k+1=256C(4n+3)6k+3+(4k+3)2k+1(XY)2k+1-8nk+8k2-16k3=256C((4n+3)3+(4k+3)(XY))(2k+1)-8nk+8k2-16k3=256CBy (XY)2k=32C1*2k+1=64kC1+1((4n+3)3+(4k+3)(XY)2k+1)(2k+1)-8nk+8k2-16k3+64kC1(2C+1)=256CBy Zm=(X+Y)2C-(4k+3)X2k+1Y2k+1((4n+3)3-Zm)(2k+1)-8nk+8k2-16k3+64kC1(2C+1)=256CBy 32+112n+48k+48n2+144nk+3*32n2+64n3=128+Zm+4n+5+256C（k=(1)）(-48k-144nk)(2k+1)-8nk+8k2-16k3+128+64kC1(2C+1)=256C8k(k-6)+8nk+64kC1(2C+1)=128+256C8k(k+2)+8nk+64kC1(2C+1)=256C （appear nk and the O of it is small.It isgoal ）(4n+3)4k+3+(4k+3)(XY)2k+1=256C16n2(4n+3)4k+1+3(8n+3)(4n+3)4k+1+(4k+3)2k+1(XY)2k+1-3*8k2-3*8k-3*16k=256C-4n*3*(4n+3)4k+1+16n2(4n+3)+32(4n+1)(4n+3)4k+1+(4k+3)2k+1(XY)2k+1-3*8k2-3*8k-3*16k=256C-4n*3*(4n+3)4k+1+16n2(4n+3)+32(4n+1)2k+1(4n+3)4k+1+(4k+3)2k+1(XY)2k+1-3*8k2-3*8k-3*16k+8nk=256CBy 8k(k+2)+8nk+64kC1(2C+1)=256C-4n*3*(4n+3)4k+1+16n2(4n+3)+32(4n+1)2k+1(4n+3)4k+1+(4k+3)2k+1(XY)2k+1-32k2-3*8k-64k=256C+64kC1(2C+1)-4n*3*(4n+3)4k+1+16n2(4n+3)+34k+2(4n+1)2k+1(4n+3)4k+1+(4k+3)2k+1(XY)2k+1-32k2-3*8k-64k+9(1-(4-1)4k)(4n+1)2k+1(4n+3)4k+1=256C+64kC1(2C+1)-4n*3*(4n+3)4k+1+16n2(4n+3)+34k+2(4n+1)2k+1(4n+3)4k+1+(4k+3)2k+1(XY)2k+1-32k2-3*8k-64k+3(16k+32k)=256C+64kC1(2C+1)-4n*3*(4n+3)4k+1+16n2(4n+3)+34k+2(4n+1)2k+1(4n+3)2k+1+(4k+3)2k+1(XY)2k+1-32k2-3*8k-16k-3*32k+34k+2(4n+1)2k+1(4n+3)2k+1((4n+3)2k-1)=256C+64kC1(2C+1)-4n*3*(4n+3)4k+1+16n2(4n+3)+34k+2(4n+1)2k+1(4n+3)2k+1+(4k+3)2k+1(XY)2k+1-32k2-3*8k-16k-3*32k+5(8*nk+8k+16k-32k2)=256C+64kC1(2C+1)-4n*3*(4n+3)4k+1+16n2(4n+3)+34k+2(4n+1)2k+1(4n+3)2k+1+(4k+3)2k+1(XY)2k+1-3*32k+5*8*nk+5*16k=256C+64kC1(2C+1)-4n*3*(4n+3)4k+1+16n2(4n+3)+(9(4n+1)(4n+3)+(4k+3)(XY))(2k+1)-3*32k+5*8*nk+5*16k=256C+64kC1(2C+1)-4n*3*(4n+3)4k+1+16n2(4n+3)+(9(4n+1)(4n+3)+(4k+3)(XY)2k+1)(2k+1)-3*32k+5*8*nk+5*16k+128kC=256CBy Zm=(X+Y)2C-(4k+3)X2k+1Y2k+1-4n*3*(4n+3)4k+1+16n2(4n+3)+(9(4n+1)(4n+3)-Zm)(2k+1)-3*32k+5*8*nk+5*16k=256CBy 32+112n+48k+48n2+144nk+3*32n2+3*64k=Zm+4n+5+256C-4n*3*(4n+3)+16n2(4n+3)+(9*4n-48k-144nk-3*64k)(2k+1)-3*32k+5*8*nk+5*16k=256C-4n*3*(4n+3)+16n2(4n+3)+(-48k-144nk)2k+9*4n=256Ck=(>=2) （use nk before it）33k(4n+3)4k+3+(4k+3)33k(XY)2k+1=1024C(4n+3)3k(4n+3)6k+2+(4k+3)3k+1(XY)3k+1+32kC+4k2(2C+1)+16k(2C+1)=1024C4n(4n+3)6k+2+3k+1(4n+3)6k+2+(4k+3)3k+1(XY)3k+1+4k2(2C+1)+16k(2C+1)+32kC=1024C4n(4n+3)2+33k+1(4n+3)6k+2+(4k+3)3k+1(XY)3k+1+32kC+4k2(2C+1)+3k(1-(4-1)2k)(4n+3)6k+2+16k(2C+1)=1024C4n(4n+3)2+(3(4n+3)2+(4k+3)(XY))(3k+1+4k(2C+1)+128C+4*3k(3k+1)(2C+1))+4k2(2C+1)+8k+16k+32kC+16k(2C+1)=1024C4n(4n+3)2+(3(4n+3)2+(4k+3)(XY)2k+1)(3k+1+4k(2C+1)+128C)+4k2(2C+1)+8k+32kC=1024CBy Zm=(X+Y)2C-(4k+3)X2k+1Y2k+14n(4n+3)2+(3(4n+3)2-Zm)(3k+1+4k(2C+1)+128C)+4k2(2C+1)+8k+32kC=1024CBy 32+112n+48k+48n2+3*32n2+144nk+64n3+64kC=Zm+4n+5+1024C （byk=(>=4)）4n(4n+3)2+((-9*4n-48k-3*32n2-144nk-64n3-64kC)(3k+1+4k(2C+1)+128C)+4k2(2C+1)+8k+32kC=1024C8k-27*4nk+32kC-48k+4k2(2C+1)=1024C(3-8)*8k+4nk+32kC+4k2(2C+1)=1024C3*8k+4nk+32kC+4k2(2C+1)=1024CWhen k=(2)By 4n+4k+8=32Cn+2=(2)n+6=(>=3)(6+n)4k+32kC+4k2(2C+1)=1024Ck=(>=3)So k=(>=6)Zm=34k+3+4n(4k+3)34k+2+16n2(4k+3)(2k+1)34k+1+(4k+3)(2k+1)(4k+1)64n334k-1+211CZm=34k+3+4n34k+3+16n234k+2+64n334k+211C=27+27*4n+9*16n2+64n3+27*16k+211C(4n+3)8k+6-(4k+3)2(XY)4k+2=211C34k(4n+3)8k+6-34k(4k+3)2(XY)4k+2=211C34k(16n2+3*8n)(4n+3)8k+4+34k+2(4n+3)8k+4-(4k+3)4k+2(XY)4k+2=211C(16n2+3*8n)(4n+3)4+(3(4n+3)2+(4k+3)(XY))((4k+2)(3(4n+3)2)4k+1-(3(4n+3)2+(4k+3)(XY))(8C+1))=211C(16n2+3*8n)(4n+3)4+(3(4n+3)2+(4k+3)(XY)2k+1)((4k+2)(3(4n+3)2)4k+1-(3(4n+3)2+(4k+3)(XY)2k+1)(8C+1))=211CBy Zm=(X+Y)2C-(4k+3)X2k+1Y2k+1(16n2+3*8n)(4n+3)4+(3(4n+3)2-Zm)((4k+2)(3(4n+3)2)4k+1-(3(4n+3)2-Zm)(8C+1))=211CBy Zm=27+27*4n+9*16n2+64n3+27*16k+211C(16n2+3*8n)(4n+3)4+(-9*4n-3*32n2-64n3-27*16k)(2(3(4n+3)2)-(-9*4n-3*32n2-64n3-27*16k)(8C+1))=211C （note：(8C+1) in it is result of operation）-81*64n2=(>=9)InvalidRefer to the introduction：The manuscript is for the most part simple.Now let's look at some of thewrong things. And it can help us find the right way.Refer to the content：Let Y is evenXN+YN=ZN X,Y,Z co-primeXN=(ZN/2+YN/2)(ZN/2-YN/2) (ZN/2+YN/2),(ZN/2-YN/2) co-primeN>2 N=m/n m,n are oddX1N=ZN/2+YN/2(X1N/2+YN/4)(X1N/2-YN/4)=ZN/2X1N/2+YN/4=Z1N/2In the same wayX2N/4+YN/16=Z2N/4…………When (a1/4+b1/4)(a1/4-b1/4)(a1/2+b1/2)=(1)a1/4-b1/4=(1/4)In addtionZ1N/2-X1N/2=(>=1/2)YN/4=(>=1/2) YN=(>=2)Z2N/4-X2N/4=(>=1/4)YN/16=(>=1/4) YN=(>=4)…………YN infiniteWhen m,n are even,In the same way.Now let's think about the following questions.Zm=(X+Y)2C-(4k+3)X2k+1Y2k+1(4n+3)4k+3+(4k+3)X2k+1Y2k+1=(X+Y)2CWe're going to use A2C+1+B2C+1=(A+B)(2C+1) note Zm=64C-4n-5In the above equation, A plus B is equal to (D), and D is large enough.By X2k+1Y2k+1 It constrains 2C+1 to be 2k+1 or 4k+1 or 6k+1 or 8k+1Notice they're both 4C+1 Rather than 4C-1When we expand parts of (4n+3)4k+3, part of (4n+3)4k+3 becomes part of A.That part of (4n+3)4k+3 is E.The other part is F.By A2C+1+B2C+1=(A+B)(4C+1)E+(F+(4k+3)XY)(2C+1)=211CE+(F-Zm)(2C+1)=211CE+F+4C(F-Zm)-Zm=211C4C(F-Zm)=211CSo It doesn't make sense to change n.because X2k+1Y2k+1 constrains 2C+1to be 2k+1 or 4k+1 or 6k+1 or 8k+1And we found that k=(1) is always true.Because,K is what we added.When we add a k, we definitely add a k2.Ork*C is too big and meaningless.The way it came about was too simple.Of course.n*k is the product of change.What is remarkable is that it ischaotic.Unconstrained.We see it in many wrong cases.Each time,it'sgoing to be an 8n*k one day, a 16n*k the next.So there has to be a situation. It takes it out of the base case.And then we need one that has k in it and it =(G) G is low number.And then we have to get out of the way.Because the original formula is not going to get us what we want.Because it's in this form,(4n+3)3+(4k+3)X2k+1Y2k+1=(X+Y)2CSo if we still have a solution to this, we can definitely solve for X and Y.

The part of Fermat's theorem

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The part of Fermat's theorem

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