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 (c−a) take out of square brackets.60

Then we obtain:

A_>iB_>iE_>i=(2m)^>n (10)

where A_>i = c−b=a−2m; B_>i=c−a=b−2m; E_>i – polynomial of power n−2.

Equation (10) is a ghost that can be seen clearly only on the assumption that the number {a^>n+b^>n−c^>n} is reduced when (1) is substituted into (8). But if it is touched at least once, it immediately crumbles to dust. For example, if A_>i×B_>i×E_>i=2m^>2×2^>n-1m^>n-2 then as one of the options could be such a system:

A_>iB_>i=2m^>2

E_>i=2^>n-1m^>n-2

In this case, as we have already established above, it follows from A_>iB_>i=2m^>2 that for any natural number m the solutions of equation (1) must be the Pythagoras’ numbers. However, for n>2 these numbers are clearly not suitable and there is no way to check any other case because in a given case (as with any other variant with the absence of solutions) another substitution will be definitely unlawful and the ghost equation (10), from which only solutions can be obtained, disappears.