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Author : Spanu Dumitru Viorel
Address : Street Marcu Mihaela Ruxandra no.5 , 061524 Romania , Bucharest
E-mail : spanuviorel@yahoo.com
spanu_ dumitruviorel@yahoo.com
Phone : +40214131107
+40723880545
A property of prime numbers .
IT DOES NOT EXIST three distinct prime numbers so that
they should satisfies the equation which represents
Pithagora`s Theorem :
X2 + Y2 = Z2 ( 1 )
Let it be X , Y , Z three natural numbers ≥ 2 .
There are no solutions in prime numbers for the equation 1 .
Sketch of proof :
Case 1 .
If X and Y are both even numbers so that
X2 + Y2 to be the square of a natural number ,
than Z2
is a even number , and this implies that
_______
√ Z2 can not be an odd number , and so it is obvious that
Z can not be a prime number .
Case 2 .
If X is an even number and Y is an odd number
let it be X = 2k and
let it be Y = 2j + 1 ;
than
X2 + Y2 = 4k2 + 4j2 + 4j + 1 .
But we stated the condition that simultaneously the three solutions of equation 1 to be prime numbers .
The only even prime number is 2 .
Than , iff X = 2 and of course X2 = 4 , that implies
X2 + Y2 = 4 + 4j2 + 4j + 1 = 4 ( j2 + j ) + 5 .
____________________
Z = √ 4 ( j2 + j ) + 5
The expression
4 ( j2 + j ) + 5
which appears under the radical
could not be a perfect square of a natural number .
___________________
Z = √ 4 j( j + 1 ) + 5 ;
j and ( j + 1 ) are consecutive numbers , so that one of them is a even number .
____________
Z = √ 8 p + 5
|
There are no perfect squares of the form 8p + 5 . Trebuie sa studiem daca enuntul de mai sus este o conjectura sau daca putem gasi o demonstratie . |
Z2 can not be for this reason a perfect square , and this
implies that Z can not be a natural number ; and so Z can
not be a prime number .
Case 3 .
This case is resembling with case 1 .
|
X2 + Y2 = Z2 Let it be both X and Y odd natural numbers . Z2 is an even number . With the above conditions , Z2 can not be a perfect square . Trebuie sa studiem daca enuntul de mai sus este o conjectura sau daca putem gasi o demonstratie . |
Even , we suppose , by reductio ad absurdum that
Z2 is a perfect square , because it is a even number it
can not be the square of an odd natural number .
So that Z can not be an odd natural number , and by
cosequence it can not be a prime number .
|
Conjectura .
It must be said that the prime number 2 can not be a solution of the equation representing Pithagora`s Theorem when the other two solutions are required to be prime numbers , because the gap between the squares of two distinct prime numbers which are both ≥ 3 is always greater than 22 = 4 . Aceast statement de mai sus este o propozitie matematica adevarata . |
Author : Spanu Dumitru Viorel
E-mail : spanuviorel@yahoo.com
Phone : +40214131107
A property of prime numbers .
IT DOES NOT EXIST three distinct prime numbers so that
they should satisfies the equation which represents
Pithagora`s Theorem :
X2 + Y2 = Z2 ( 1 )
Let it be X , Y , Z three natural numbers ≥ 2 .
There are no solutions in prime numbers for the equation 1 .
Sketch of proof :
Case 1 .
If X and Y are both even numbers so that
X2 + Y2 to be the square of a natural number ,
than Z2
is a even number , and this implies that
_______
√ Z2 can not be an odd number , and so it is obvious that
Z can not be a prime number .
Case 2 .
If X is an even number and Y is an odd number
let it be X = 2k and
let it be Y = 2j + 1 ;
than
X2 + Y2 = 4k2 + 4j2 + 4j + 1 .
But we stated the condition that simultaneously the three solutions of equation 1 to be prime numbers .
The only even prime number is 2 .
Than , iff X = 2 and of course X2 = 4 , that implies
X2 + Y2 = 4 + 4j2 + 4j + 1 = 4 ( j2 + j ) + 5 .
____________________
Z = √ 4 ( j2 + j ) + 5
The expression
4 ( j2 + j ) + 5
which appears under the radical
could not be a perfect square of a natural number .
___________________
Z = √ 4 j( j + 1 ) + 5 ;
j and ( j + 1 ) are consecutive numbers , so that one of them is a even number .
____________
Z = √ 8 p + 5
|
There are no perfect squares of the form 8p + 5 .
Trebuie sa studiem daca enuntul de mai sus este o conjectura sau daca putem gasi o demonstratie . |
Z2 can not be for this reason a perfect square , and this
implies that Z can not be a natural number ; and so Z can
not be a prime number .
Case 3 .
This case is resembling with case 1 .
|
X2 + Y2 = Z2 Let it be both X and Y odd natural numbers .
Z2 is an even number . With the above conditions , Z2 can not be a perfect square . X2 + Y2 = Z2 |
Even , we suppose , by reductio ad absurdum that
Z2 is a perfect square , because it is a even number it
can not be the square of an odd natural number .
So that Z can not be an odd natural number , and by
cosequence it can not be a prime number .
|
It must be said that the prime number 2 can not be a solution of the equation representing Pithagora`s Theorem when the other two solutions are required to be prime numbers , because the gap between the squares of two distinct prime numbers which are both ≥ 3 is always greater than 22 = 4 .
Aceast statement de mai sus este o propozitie matematica adevarata . |
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