Proof-6
There is a wonderful proof for this theorem, based on the
characteristics of the odd and even cubes and its corresponding
cube roots. The cubes of a number is always in the form of
7p or 7p± 1
First few numbers whose cubes are in the form of 7p and 7p± 1
are tabulated below
7p+1 7p 7p-1
1, 2, 4 7,14,21 3, 5, 6
8, 9,11 28,35,42 10,12,13
15,16,18 49,56,63 17,19,20
22,23,25 70,77,84 24,26,27
29,30,32 91,98,105 31,33,34
… … …
The general forms of cube roots under different categories are
[ (1+7p’) [(3+7p’)
(2+7p’) [7(p’+1)] (5+7p’)
(4+7p’)] (6+7p’)]
where p’ = 0,1,2,3…..
It implies some conditions for the choice of the root numbers
to form a cubical relation a^3+b^3 = c^3. The cubes of
a,b,c cannot be in the same form of 7p.Because the
cubical relation will then be reducible by 7.If they are all
either in the form 7p-1 or 7p+1 ,then the sticking number
-2 or +2 in one side cannot be equated with -1 or +1 in the
other side.
If c^3 is in the form of 7p±1,then a^3 and b^3 must be in
the form of 7p and 7p±1 respectively or vice-versa. If c^3
is in the form of 7p, then a^3 and b^3 must be in the form
of 7p+1 and 7p-1 respectively or vice-versa.
In all the cases, the condition becomes,
p(a) +p(b) = p(c)
i.e., the integral number of 7 in c^3 must be equal to the sum
of integral numbers of 7 in a^3 and b^3.
If the root number (a,b,c) is odd, p will be even if its cube
is in the form of either 7p+1 or 7p-1,and p will be odd if its
cube is in the form of 7p.
If the root number is even, p will be odd if its cube is in the
form of 7p+1 or 7p-1, and p will be even if the cube is in the
form of 7p.
It is found that the ending digit of p is also closely associated
with the ending digit of the root number . However it depends
upon the form of the cubes represented in terms of multiples of 7.
Ending digit of p ____________________________________________
Ending digit of the root number
form 0 1 2 3 4 5 6 7 8 9
7p+1 7 0 1 8 9 2 5 6 3 4
7p 0 3 4 1 2 5 8 9 6 7
7p-1 3 6 7 4 5 8 1 2 9 0
The root number with a given ending digit whose cube is in
a given form cannot have all possible values, but some
selective values which are repeating with a periodicity of 70.
Root number ending with
0 1 2 3 4 5 6 7 8 9
30 1 2 23 4 15 16 37 8 9
50 11 22 43 44 25 36 57 18 29
60 51 32 53 64 65 46 67 58 39
7p+1 -------------------------------------------------------------
100 71 72 93 74 85 86 107 78 79
120 81 92 113 114 95 106 127 88 99
130 121 102 123 134 135 116 137 128 109
10 31 12 3 24 5 6 17 38 19
20 41 52 13 34 45 26 27 48 59
40 61 62 33 54 55 66 47 68 69
7q-1 ------------------------------------------------------------------------------------------ 80 101 82 73 94 75 76 87 108 89
90 111 122 83 104 115 96 97 118 129
110 131 132 103 124 125 136 117 138 139
70 21 42 63 14 35 56 7 28 49
7r ------------------------------------------------------------------------------------------
140 91 112 133 84 105 126 77 98 119
The ideal mathematical relation, a^3 + b^3 = c^3 to be true,
1. If a^3 is in the form of 7p(a)+1 then b^3 must be in the form of 7p(b)-1 or vice-versa
in order to keep c^3 to be in the form 7p
2. If a and b are odd then c must be even. If all of them are even, it will be reducible
And all of them cannot be odd as the sum of two odd numbers cannot be odd.
3. Both a and b should always be less than c.
4. However the cubical relation satisfies a unique condition a+b = c+6m that is a+b >c.
5. Since the sum of two cubes is equal to a cube, the ending digits of the root numbers
must have an inherent correlation.
All these requirements are fulfilled for all choices of numbers to a,b and c..For example,
if a ends with 1 and b also ends with 1, then c will end with 8. The suitable number for a
whose cube is in the form of 7p+1 are 1,11, and 51 ,These numbers are cyclically
varying after 70..Hence the general form of these numbers are (1+70p’).(11+70p’),
(51 + 70p’). The suitable number for b whose cube is in the form of 7p-1 are 31,41,
and 61 with general form (31+70p’ ),(41+70p’) and (61+70p’).The even numbers
ending with 8 and divisible by 7 (as its cube is divisible by 7) take the general form
(28 +70p’).
The requirement (4) demands that
a+b = c+ 6m
(10x +l) + (10y+1) = (10z+8) +6m or 10(x+y) = 10z +6(m+1)
Or (m+1) must be in the form of 5k
x+y = z +3k
The possible values of x and y are (0,1 and 5), (3,4 and 6 ) respectively with or
without adding some multiples of 7 with them.The corresponding possible values
of z are 2,9,16……..In any case x and y cannot be greater than z. When we
apply all these conditions together, no set of allowed numbers will satisfy the
numeral relation.Hence we can conclude that this kind of relation is not at all
existing.
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