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The Wonders of Arithmetic from Pierre Simon de Fermat
Год написания книги
2021
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)?[a
?(2m)
]=(2m)
(9)
In this identity natural numbers a, b, c, n, m of course, may be any. The only question is whether there are such among them that {a
+b
?c
} will be zero? However, the analogy with the solution of the Pythagoras’ equation ends on this since the substitution of (1) in (8) is not substantiated in any way. Indeed, by substituting (1) in (3), it is well known that the Pythagoras’ equation has as much as you like solutions in natural numbers, but for cases n>2 there is no single such fact. Therefore, the substitution of the non-existent equation (1) in (8) is not excluded, what should lead to contradictions. Nevertheless, such a substitution is easily feasible and as a result we obtain an equation very similar to (4), which gives solutions to the Pythagoras equation. Taking into account this circumstance, we yet substitute (1) in (8) as a test, but at the same time modify (8) so, that factor (c?a) take out of square brackets.[60 - Taking into account that c?a=b?2m the expression in square brackets of equation (8) can be transformed as follows: (c++b)
? (a++2m)
= с
? a
+ c
b? a
2m+ c
b
? a
(2m)
+ … +b
? (2m)
; с
? a
= (с ? a)(c++a)
; c
b ? a
2m = 2m(c
? a
) + c
(b ? 2m) = (c ? a)[2m(c++a)
+ c
]; c
b
?a
(2m)
= (2m)
(c
? a
) + c
(b
? 4m
) = (c ? a)[4m
(c++a)
+ c
(b +2m)]; b
? (2m)
= (b ? 2m)(b++2m)
= (c ? a)(b++2m)
All differences of numbers except the first and last, can be set in general form: c
b
? a
(2m)
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