A new method for proving Fermat theorem

Fermat theorem: Xn+yn=zn (n>=3), and X, Y, z are both positive integers, the equation is not established.

12= 1 13 = 1 14 = 1
22= 4 23 = 8 24 = 16
32= 9 33 = 27 34 = 81
42= 16 43 = 64 44 = 256
52= 25 53 = 125 54 = 625
62= 36 63 = 216 64 = 1296
72= 49 73 = 343 74 = 2401
82= 64 83 = 512 84 = 4096
92= 81 93 = 729 94 = 6561
102=100 103=1000 104=10000
112=121 113=1331 114=14641
122=144 123=1728 124=20736
132=169 133=2197 134=28561
142=196 143=2744 144=38416
152=225 153=3375 154=50625

13= 1 14 = 1 15=1
23= 8 24 = 16 25=32
33= 27 34 = 81 35=243
43= 64 44 = 256 45=1024
53= 125 54 = 625 55=3125
63= 216 64 = 1296 65=7776
73= 343 74 = 2401 75=16807
83= 512 84 = 4096 85=32768
93= 729 94 = 6561 95=59049
103=1000 104=10000 105=100000
113=1331 114=14641 115=161051
123=1728 124=20736 125=248832
133=2197 134=28561 135=371293
143=2744 144=38416 145=537824
153=3375 154=50625 155=759375
165=1048576
175=1419857
185=1889568


Prove:

We first prove that the time of n=2, Z first must be greater than or equal to 3 of the number.

When you take z=3, you will find 1 2=3 2 is not established, but 1 2=5, and 5 is less than 3 2, that is 9, and the difference is 4, when z equals 4 o'clock, also have to clear an idea, if you need Z squared is the other two number of square sum, then, these two numbers must have a greater than Z 2/ 2, and the other is less than Z 2/2. Now say that when Z equals 4 o'clock, 2 2+3 2=13, or less than Z squared, but the difference becomes 3, that is, the two equations are approaching. When z equals 5,3 2+4 2=25,ok, exactly equal to 5 squared, if we do not guess correctly, then X2 and Y2 will be greater than the square of Z. The fact is, however, you will find this value getting bigger, look at this 13 2 half 84.5, closest to it is 10 2 and 9 2, but their and is 181, to be greater than 169, but 8 2 and 11 2 is 185, greater than 181, which means that the nearest Z 2/ 2 of the two number of squares is the closest to Z 2, in 13 2 and 14 2 o'clock, you will find that their two Z 2/2 is the same, which means that when z is increasing, there will be more Z at the same time the sum of the squares of two adjacent numbers, then you can move the number less than Z 2/2 to a small position, In order to better fit Z 2, which means that it can be combined, because it has the premise that can be combined, there are many combinations, then return to the point, prove x3+y3=z3.

First z needs to be greater than or equal to 5, because only 5 of the time, Z 3/2 has a big small two numbers and, but, with the increase of Z 3/2, the sum of squares and Z 2 difference will increase, at 12 3 will appear a drop, because when Z is increasing, There will be more Z at the same time between the sum of the squares of two adjacent numbers, resulting in a smaller distance difference, and 12 3 is the smallest first difference of 1, the next can be divided into two cases, one is normal increase, that is, Z increase 1,x and y simultaneously increase 1, the other is Z increase 1 but X and Y will not increase, But they will be suddenly close to the second side of Z, but, not less than the previous sudden fall distance Z3, and Z4 is so proved that the increase in power, only the X and y of the n-squared and Z of the n-th variance value changes more and more quickly only, and, it is never possible to equal z of the n-th square.

The certificate is completed.

A new method for proving Fermat theorem