The Wonders of Arithmetic from Pierre Simon de Fermat - страница 51

a+b=c+2m (2)

where m is a natural number.

To obtain formula (2) we note that a≠b since otherwise 2a^>n=c^>n what is obviously impossible. Consequently, a>n-1+b^>n-1)>c^>n-1 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 even56.

At first, we verify the effectiveness of the method for the case n = 2 or the Pythagoras’ equation a^>2+b^>2=c^>2. 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^>2+b^>2−c^>2}+2(c−b)(c−a)=4m^>2 (3)

Substituting the Pythagoras’ equation in (3), we obtain:

A_>iB_>i=2m^>2 (4)

where taking into account the formula (2):

A_>i=c−b=a−2m; B_>i=c−a=b−2m (5)

Now we decompose the number 2m^>2 into prime factors to get all the A_>iB_>i options. For primes m there are always only three options: 1×2m^>2=2×m^>2=m×2m. In this case A_>1=1; B_>1=2m^>2; A_>2=2; B_>2=m^>2; A_>3=m; B_>3=2m. Since from (5) it follows a=A_>i+2m; b=B_>i+2m; and from (2) c=a+b−2m; then we end up with three solutions:

1. a_>1=2m+1; b_>1=2m(m+1); c_>1=2m(m+1)+1

2. a_>2=2(m+1); b_>2=m(m+2); c_>2= m(m+2)+2 (6)

3. a_>3=3m b_>3=4m; c_>3=5m

Equations (6) are the solutions of the Pythagoras’ equation for any natural number m. If the number m is composite, then the number of solutions increases accordingly. In particular, if m consists of two prime factors, then the number of solutions increases to nine57. Thus, we have a new way of calculating all without exception triples of Pythagoras’ numbers, while setting only one number m instead of two numbers that must be specified in the Pythagoreans identity. However, the usefulness of this method is not limited only to this since the same key formula (2) is also valid for obtaining a general solution of equations with higher powers.

Using the method to obtain solutions of (1) for the case n=2, it is also possible to obtain solutions for n>2 by performing the substitution (1) in (2) and exponentiating n both sides of (2). To do this, first we derive the following formula58:

(x+y)^>n=z^>n=zz^>n-1=(x+y)z^>n-1=xzz^>n-2+yz^>n-1=

x(x+y)z^>n-2+yz^>n-1=x^>2zz^>n-3+y(z^>n-1+xz^>n-2)+…

(x±y)^>n=z^>n=x^>n±y(x^>n-1+x^>n-2z+x^>n-3z^>2+…+xz^>n-2+z^>n-1) (7)

We will name the expression in brackets consisting of n summand a symmetric polynomial and we will present it in the form (x ++ z)_>n as an abridged spelling. Now using formula (7), we will exponentiating n both sides of formula (2) as follows.

[a−(c−b)]^>n=a^>n+{b^>n−c^>n+(c^>n−b^>n)}−(c−b)[a^>n-1+a^>n-22m+…

+ a(2m)^>n-1+(2m)^>n-1]=(2m)^>n

Now through identity

(c^>n−b^>n)=(c−b)(c^>n-1+c^>n-2b+…+cb^>n-2+b^>n-1) we obtain:

{a^>n+b^>n−c^>n}+(c−b)[(c++b)_>n−(a++2m)_>n]=(2m)^>n (8)

Equation (8) is a formula (2) raised to the power n what can be seen after substituting c−b=a−2m in (8) and obtaining the identity59:

{a^>n+b^>n−c^>n}+(c^>n−b^>n)−[a^>n−(2m)^>n]=(2m)^>n (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^>n+b^>n−c^>n} 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 (