prime factors of Pythagorean hypotenuses

prime factors of Pythagorean hypotenuses

The possible

hypotenuses

of the

Pythagorean triangles

(

http://

planetmath

.org/PythagoreanTriangle

) form the

infinite

sequence

5

,

10

,

13

,

15

,

17

,

20

,

25

,

26

,

29

,

30

,

34

,

35

,

37

,

39

,

40

,

41

,

45

,

…

the mark of which is

http://oeis.org/search?q=a009003&

language

=english&go=Search

A009003 in the corpus of the

integer
sequences

of

http://oeis.org/

OEIS.  This sequence has the

subsequence

A002144

5

,

13

,

17

,

29

,

37

,

41

,

53

,

61

,

73

,

89

,

97

,

101

,

109

,

113

,

137

,

…

of the odd

Pythagorean primes

.

Generally, the hypotenuse

c

of a Pythagorean triangle
(

Pythagorean triple

) may be characterised by being the

contraharmonic mean

c

=

u

2

+

v

2

u

+

v

of some two different

integers

u

and

v

(as has been shown
in the parent entry), but also by the

Theorem

.

A

positive integer

c

is the length of the
hypotenuse of a Pythagorean triangle if and only if at least
one of the

prime factors

of

c

is of the form

4

⁢

n

+

1

.

Lemma 1.

All prime factors of the hypotenuse

c

in
a

primitive Pythagorean triple

are of the form

4

⁢

n

+

1

.

This can be proved here by making the antithesis that there
exists a prime

4

⁢

n

-

1

dividing

c

.  Then also

4

n

-

1

∣

c

2

=

a

2

+

b

2

=

(

a

+

i

b

)

(

a

-

i

b

)

where

a

and

b

are the

catheti

in the triple.  But

4

⁢

n

-

1

is prime also in the ring

ℤ

⁢

[

i

]

of the

Gaussian
integers

, whence it must divide at least one of the factors

a

+

i

⁢

b

and

a

-

i

⁢

b

.  Apparently, that would imply that

4

⁢

n

-

1

divides both

a

and

b

.  This means that the
triple

(

a

,

b

,

c

)

were not

primitive

, whence
the antithesis is wrong and the lemma true.

□

Also the

converse

is true in the following form:

Lemma 2.

If all prime factors of a positive integer

c

are of the form

4

⁢

n

+

1

, then

c

is the hypotenuse in
a Pythagorean triple.  (Especially, any prime

4

⁢

n

+

1

is
found as the hypotenuse in a primitive Pythagorean triple.)

Proof

.  For proving this, one can start from Fermat’s
theorem, by which the

prime numbers

of such form are

sums of
two squares

(see the

http://en.

wikipedia

.org/wiki/Proofs_of_Fermat's_theorem_on_sums_of_two_squares

Theorem on sums of two squares by Fermat

).
Since the sums of two squares form a set

closed under

multiplication

, now also the

product

c

is a sum of two
squares, and similarly is

c

2

, i.e.

c

is the hypotenuse
in a Pythagorean triple.

□

Proof of the Theorem.

Suppose that

c

is the
hypotenuse of a Pythagorean triple

(

a

,

b

,

c

)

; dividing the
triple members by their

greatest common factor

we get a
primitive triple

(

a

′

,

b

′

,

c

′

)

where

c

′

∣

c

.  By
Lemma 1, the prime factors of

c

′

, being also prime factors of

c

, are of the form

4

⁢

n

+

1

.

On the contrary, let’s suppose that a prime factor

p

of

c

=

p

⁢

d

is of the form

4

⁢

n

+

1

.  Then Lemma 2 guarantees
a Pythagorean triple

(

r

,

s

,

p

)

, whence also

(

r

⁢

d

,

s

⁢

d

,

c

)

is
Pythagorean and

c

thus a hypotenuse.

□

Title

prime factors of Pythagorean hypotenuses

Canonical name

PrimeFactorsOfPythagoreanHypotenuses

Date of creation

2014-01-31 10:48:20

Last modified on

2014-01-31 10:48:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11D09

Classification

msc 51M05

Classification

msc 11E25

Generated on Fri Feb 9 15:31:40 2018 by

LaTeXML