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The Wonders of Arithmetic from Pierre Simon de Fermat
Год написания книги
2021
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a
+b
=c
(1)
To prove this statement, suppose that a, b, c satisfying to (1) exist and then based on this, we can get the all without exception solutions to this equation in general form. To this aim we use the key formula method, in which one more equation is added to the initial equation so that it becomes possible to obtain solution (1) in a system of two equations. In our case the key formula is:
a+b=c+2m (2)
where m is a natural number.
To obtain formula (2) we note that a?b since otherwise 2a
=c
what is obviously impossible. Consequently, a<b<c and we can state that (a
+b
)>c
whence (a+b)>c. Since in (1) cases with three odd a, b, c, as well as one odd and two even are impossible, the numbers a, b, c can be either all even or two odd and one even. Then from (a + b) > c follows formula (2) where the number 2m is even[56 - Fermat discovered formula (2) after transforming the Pythagoras’ equation into an algebraic quadratic equation – see Appendix IV story Year 1652. However, an algebraic solution does not give an understanding the essence of the resulting formula. This method was first published in 2008 [30].].
At first, we verify the effectiveness of the method for the case n = 2 or the Pythagoras’ equation a
+b
=c
. Here the key formula (2) applies and you can get a solution to the system of equations (1), (2) if you substitute one into another. To simplify it, we will square both sides of (2) to make the numbers in (1) and (2) proportionate. Then (2) takes the form:
{a
+b
?c
}+2(c?b)(c?a)=4m
(3)
Substituting the Pythagoras’ equation in (3), we obtain:
A
B
=2m
(4)
where taking into account the formula (2):
A
=c?b=a?2m; B
=c?a=b?2m (5)
Now we decompose the number 2m
into prime factors to get all the A
B
options. For primes m there are always only three options: 1?2m
=2?m
=m?2m. In this case A
=1; B
=2m
; A
=2; B
=m
; A
=m; B
=2m. Since from (5) it follows a=A
+2m; b=B
+2m; and from (2) c=a+b?2m; then we end up with three solutions:
1. a
=2m+1; b
=2m(m+1); c
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