Arturo Reghini here extends the Pythagorean arithmetic to those properties which actually exist. The end of the third chapter of “Sacred Numbers” presents some Pythagora classics: Egyptian triangles and a hypotenuse theorem extension.

My translation from “I numeri Sacri nella Tradizione Pitagorica Massonica” or sacred numbers in masonic Pythagorean tradition. The second part of the third chapter is on the dryad of odd prime numbers within the decade.

A second way to reach the number five is instead suggested by Plutarch’s consideration. The mention of Plutarch is located in De Iside et Osiris and reconnects to the “Egyptian” triangle rectangle, the simplest of integers 3, 4, and 5 triangles. Geometrically the Pythagorean theorem, which applies to every right triangle, says that in a right triangle the sum of the squares created on catheti is equal to the square on the hypotenuse, arithmetically, when the sides of the triangle are integers, it happens that the sum of the squares of these integers is equal to the square that has a side to the hypotenuse. In the case of the Egyptian triangle where the short sides are the three and four and five is the hypotenuse, Plutarch presents an analogical interpretation of the Pythagorean theorem to the effect that the five would be the result or fruit of the spiritual action of the three vertically set and symbolizing the male above the horizontal base of the four that symbolizes the female. Thus, the five would come not from integers but from polygonal numbers and precisely from square numbers.

The triad of consecutive numbers 3, 4, 5, therefore, displays the property that the sum of the squares of the first two is equal to the square of the third. Indeed, it is easy to recognize that this is the only triad of three consecutive integers to have this property; in fact, denoting by x – 1, x and x + the three consecutive numbers in the equation

(x -1)² + x² = ( x + 1)²

that is x² – 4× = 0 has only the solutions x = 0 and x = 4. If then instead of squares are considered three consecutive triangular or pentagonal or pentagonal of the same gender r, and one tries, when it happens, that the sum of the first two polygons is equal to the third, is that this fact occurs only when the polygons are squares, one finds that this occurs only when polygons are square and namely in the case of the third, fourth and fifth square.

In fact, the polygonal x ° gender r is expressed by the formula

and consider the indeterminate equation in the unknowns r and x

P (r , x – 1) + P (r , x) = P (r , x + 1).

It admits no solution other than the solution r = 4, x = 4.

In fact, by replacing symbols their expressions, the equation becomes:

that developing and reducing becomes:

(r – 2) x² – 4 (r – 2) x + r – 4 = 0

and applying the well-known terminate formula of the quadratic equation is obtained

that is

where the discriminant is equal to 5 for r = 3, is equal to 4 for r = 4, and is always between 3 and 4 for any other value of r. We have only a rational entire and positive value of x in r = 4 and is x = 4. Therefore, as in the case of the linear numbers, the only triad of three linear consecutive numbers for which happens that the sum of the first two is equal to the third consists of the triad 1, 2, 3, so in the case of polygonal numbers the same triad of consecutive polygonal numbers of the same kind for which happens that the sum of the first two is equal to the third is constituted by the third, fourth and fifth square that’s to say by the sides of the Egyptian triangle. The Egyptian triangle is presented in this respect as a hypostasis of the fundamental triad 1, 2, 3. The triad of the numbers 3, 4, and 5 takes place in the surface manifestation or epiphany of what happens in the linear irradiation for the triad 1, 2, 3. The number five takes third place and replaces the three, as the pentagram or blazing star takes the place of the Delta, or luminous triangle, passing from the first-degree Chamber to that of the second degree.

The three numbers from this triad 3, 4, and 5 are the numbers of the sides of the Egyptian triangle. But it can be shown more generally by the following properties: In a right triangle coprime integers always happen: 1) a cathetus is even and the other two sides are odd; 2) The even cathetus is always a multiple of four; 3) The hypotenuse is always the sum of two squares, one even and the other odd, and therefore is of the form 4 n + 1; 4) The hypotenuse is never multiple of three; 5) One of the short sides is always a multiple of three. 6) One of the catheti is always a multiple of five. 7) The perimeter and the area is equal to a multiple of six.

These simple and interesting properties of right triangles in integers can be demonstrated in various ways, but since it is not easy to find these demonstrations together we will give a demonstration that the reader less demanding and mistrustful can jump to.

Let’s give first of all the general formulas for right triangles in integers prime among them. Denoting by x, y, and z the catheti of the hypotenuse, we have:

y2 =z2 –x2 =(z–x)(z+x)

and the two factors of the second member must both be squares or contain a common factor α. In this second case their sum 2 z and their difference 2 x must have in common this factor α, and as x and z are for the hypothesis prime between them, so 2 x and z 2 can not have another common factor than the one or the two.

It will therefore be:

*z *+ *x *= * α m2 z *– *x *= * α n2 *con * α *= 1 , 2

and thus

whence

y = *α m n*

and canceling the common factor α and multiplying by two the three last equalities are obtained for x, y, and z the formulas

y=2mn z=m² +n² x=m² –n²

where m and n are numbers of different parity, that is, one even and the other odd, otherwise the three sides would be equal and therefore not prime between them.

The even cathetus y result therefore multiple of 4, the hypotenuse and the other cathetus are odd, and so the perimeter is even and the area is even because it is given by the semi-product of catheti.

examples:

m=2,n=1 ; x= 4, y=3,z=5

m=3,n=2 ; x=12,y=5,z=13

The general formula that gives the hypotenuse z = m² + n² shows that it is always equal to the sum of two squares, one even and the other odd, and therefore is always of the form 4 p + 1. The cathetus odd, if m is even and n odd, is of the form 4p +1, because for m = 2 h +1 and n = 2 k one has: x = m² – n² = 4 h² + 4 h + 1 – 4 k², while if it happens the vice versa is of the form 4 q – 1. It can be shown vice versa (Fermat) that there is always a right triangle in integer numbers which has as its hypotenuse a prime of the form 4 n + 1, and more right triangles in integer numbers, if the hypotenuse is a product of the primes of this form.

We demonstrate that one of the sides is always a multiple of 5.

In fact, if none of the catheti is a multiple of five they have the shapes

x = 5 h ± 1 or x = 5 h ± 2 y = 5 k ± 1 or y = 5 k ± 2

but they can not both be of the same shape, because, as it is easy to calculate, the square of the hypotenuse should end for 2, 3, 7, 8, which is impossible, and then the sum of their squares that’s to say the square of the hypotenuse ends for five, and then the same hypotenuse ends for five that’s to say is a multiple of five.

We show that the hypotenuse can not be a multiple of three.

In fact, if for absurd the hypotenuse were a multiple of three the catheti could not be and would be then of the form x = 3 h ± 1, y = 3 k ± 1, and then the sum of their squares would be a multiple of three plus two and could not be equal to the square of the hypotenuse. The hypotenuse is therefore of the form 3 h ± 1.

We show that one of the catheti is a multiple of three.

In fact, if one of the catheti, for example, x is not a multiple of three, it would be:

y2 =z2 –x2 =(z+x)(z–x)

and being

z = 3 h ± l and x = 2 k ± l

it happens that in all four possible cases one of the two factors to the second member results multiple of three, and then the other cathetus is a multiple of three.

To summarize: The hypotenuse and a cathetus are odd, the other is a multiple of four, one of the catheti is a multiple of three and the hypotenuse is not; one of the three sides is a multiple of five, the hypotenuse is of form 4 n + 1 and is the sum of two squares, the perimeter and the area are even.

If the hypotenuse is not a multiple of five it can be that one of the catheti is a multiple of 3 and 5 and the other of 4, for example in the triangle (8, 15, 17), or that one of the catheti is a multiple of 3 and 4 and the other of 5 for example in the triangle (5, 12, 13), or in the end that one of the catheti is an as a multiple of 3 so that of both 4 and 5, for example in the triangle (60, 11, 61). If the hypotenuse is a multiple of five can be that one of the catheti is a multiple of 3 and the other of 4 as in the Egyptian triangle (3, ) considered by Plutarch; or that one of the catheti is a multiple of 3 such as of 4, as for example for the triangles (33, 56, 65), (63, l6, 65), (44, 117, 125). Other cases in which only one cathetus is a multiple of 3, 4, 5 are (119, 120, 169), (120, 391, 409) …

In the case of the Egyptian triangle (3, 4, 5) the radius of the inscribed circle is 1, the diameter is 2, the cathetus and the hypotenuse are 3, 4, 5, the area is 6, the sum of the catheti is 7, the sum of a cathetus and the hypotenuse 8 and 9 and the perimeter is l2. The Egyptian triangle was used by the Egyptians to draw a right angle. Taking a string divided into three parts of their lengths 3, 4, and 5, and fixing on the ground the distant points 5, holding the other two parts and reunifying the ends, you get the Egyptian triangle and then the right angle. The square, which is a characteristic tool of the fellow freemason and serves to square the rough stone, that’s to say the particular job of the mate, is so linked to the number five, which always appears on one of the sides of a right triangle and the numbers 3 and 4 which appear always in one of two catheti.

The question that we have solved for polygonal numbers is presented for pyramidal numbers too, that is for the most important numbers of the space considered by the Pythagoreans. It is to examine whether there exists a set of three pyramidal consecutive numbers of the same kind such that the sum of the first two is equal to the third. This means we need to solve the equation:

F (x . y – 1) + F (x , y) = F (x . y + 1)

that is

the equation that after the development and reductions become:

(y3 – 6 y2 – y) x = 2 y3 – 15 y2 + 7 y + 6

whose solutions are given by

3 (y2 – 3 y – 2)

Giving y values 1, 2, 3, and 4 … we obtain the following pairs of solutions:

and it is seen that the only integer solution is given by the pair x = 10, and y = 6. The problem, therefore, admits the only solution

F (10, 5) + F (10, 6) = F (10, 7)

that is

175 + 301 = 476

it then has the properties: The single triad of three consecutive pyramidal numbers of the same kind, such that the sum of the first two is equal to the third, is constituted by the fifth, sixth, and seventh pyramidal with a decagonal base.

As the triad of numbers (3, 4, 3) solved the problem in the plane through the right triangle that had those three numbers for the sides, the list of possible numbers (5, 6, 7) solves a similar problem in space by means of the fifth, sixth and seventh pyramidal with a decagonal based. And as the triad (1, 2, 3) of the linear numbers, that solves the problem of the three consecutive integers wherein the sum of the first two is equal to the third, gives the number three, and the triad (3, 4, 5) ), which solves the problem in the field of pyramidal numbers, gives the number 7 as result of the action of five on six so to use the same language as Plutarch; We also observe that the three pyramidal with a decagonal base that solves the problem, namely the numbers 175, 301 and 476, are three multiples of seven. The sum of the three sets 1, 2, 3, 3, 4, 5, and 5, 6, 7 is equal to 36. We observe in the end that the sum of the number of diagonals of the pentagon and hexagon is equal to the number of diagonals of the heptagon, that’s to say 5 + 9 = 14, and this is the only case in which this event occurs for three consecutive polygons, that’s to say also the problem of determining three polygons having for a number of sides three consecutive integers admits a unique solution given by the triad (5, 6, 7),

The first three odd primes, that’s to say 3, 5, 7, represent the only solution to the same problem for linear numbers, polygons of the same kind, and pyramids of the same kind. It is worth observing, moreover, that the solution has for the polygon of the fourth kind, that’s to say with the squares and for pyramidal has the pyramidal of the tenth kind that’s to say with the pyramidal numbers with a decagonal base. Four and ten, are the two numbers that identify the mystery of Luciano’s Tetractys.

We believe that so far no one has demonstrated the properties set forth above on the triad of consecutive polygonal of the same kind and the similar property on consecutive pyramidal of the same kind. It is not our intention to state that the development and extension of the Pythagorean theorem for a dryad of consecutive numbers to the case of polygonal and pyramidal had already been done by the ancient Pythagoreans, but we will not even exclude such a possibility.

But keeping to the spirit of Pythagorean arithmetic and following the procedures we have come to the results that we have presented, and these properties actually exist. We have not done anything but bring them forth and bring them to the consideration of the reader to who we leave the task to assess the importance and to draw the consequences.

We add that it could be shown that the analogous problem for numbers in hyper pyramidal in the hyperspaces does not admit a solution. From the modern point of view, there are within the decade four prime numbers: 2, 3, 5, and 7, and we will meet this Tetractys, handling of regular polyhedra.

In addition, there are four numbers by the decade with the first ten and they are: 1, 3, 7, 9; indicating with the symbol of Gauss φ (n) the number of prime numbers with n and less than n (the units included) has: (10) = 4, the relationship between 4 and 10.

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