A Property _ Of Prime Numbers

Author : Spanu Dumitru Viorel Sreet Marcu Mihaela Ruxandra no. 5 , 061524 , Bucharest , Romania E-mai : [email protected] Phone : +40214131107

A property of prime numbers . IT DOES NOT EXIST three distinct prime numbers so that they would satisfies the equation which represents the Pithagorean 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 than

to be the square of a natural number

,

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 ; ( both k and j are natural numbers ) than X2 + Y2 = 4k2 + 4j2

+ 4j

+ 1

.

But we stated the condition that simultaneous 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 = √ 8p + 5 ( p is a natural number ) There are no perfect squares of the form

8p + 5 .

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 . Let it be both X and Y Z2

odd natural numbers .

is an even number .

With the above conditions , Z2 can not be a perfect square .

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 the Pithagorean 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 .