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 ttat’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)
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 the 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 end that one of the catheti is a 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, 5, and fixed 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 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 in 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)
equation that after the development and reductions becomes:
(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, 4 … we obtain the following pairs of solutions:
and it is seen that the only integer solution is given by the pair x = 10, y = 6. The problem therefore admits the only solution
F (10, 5) + F (10, 6) = F (10, 7)
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, so the list of possible numbers (5, 6, 7) solves the 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 decagonal base that solve 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 end that the sum of the number of diagonals of the pentagon and hexagon is equal to the number of diagonals of 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 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 of the same problem for linear numbers, for the polygons of the same kind and for the pyramids of the same kind. It is worth to observe, 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, the two numbers that identifies the mystery of Luciano’s Tetractys.
We believe that so far no one has demonstrated the properties set forth above on the triad of consecutives polygonal of a same kind and the similar property on consecutives pyramidal of the same kind. It is not our intention to state that the development and extension of the Pythagorean theorem for a driad 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 possibility.
But keeping to the spirit of the 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 hyperpiramidal in the hyperspaces does not admit solution. From the modern point of view there are within the decade four prime numbers: 2, 3, 5, 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) is has: (10) = 4, the relationship between 4 and 10.
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