proof of Pythagorean triplet

proof of Pythagorean triplet

Consider the

4

⁢

A

⁢

B

=

(

A

+

B

)

2

-

(

A

-

B

)

2

. Assume that

A

,

B

are

coprime

. (This is no : If

d

is the

greatest common divisor

of

A

,

B

, then one can write

A

=

d

⁢

A

′

,

B

=

d

⁢

B

′

to get

4

⁢

d

2

⁢

(

A

′

⁢

B

′

)

2

=

d

2

⁢

(

(

A

′

+

B

′

)

2

-

(

A

′

-

B

′

)

2

)

and cancel

d

2

.)

For

4

⁢

A

⁢

B

,

A

+

B

,

A

-

B

to form a

Pythagorean triple

, each of

A

,

B

must be squares. So

A

=

m

2

,

B

=

n

2

where

m

,

n

are coprime,

n

<

m

. So we have

a

=

2

⁢

m

⁢

n

,

b

=

m

2

-

n

2

,

c

=

m

2

+

n

2

(1)

and

{

a

,

b

,

c

}

are a Pythagorean triple. But this needn’t be

primitive

: If

m

,

n

are odd, then

2

∣

m

2

±

n

2

, so not all of

a

,

b

,

c

are relatively prime.

Suppose

2

⁢

m

⁢

n

,

m

2

-

n

2

,

m

2

+

n

2

are

pairwise coprime

. Then

gcd

⁡

(

2

⁢

m

⁢

n

,

m

2

+

n

2

)

=

gcd

⁡

(

2

⁢

m

⁢

n

,

(

m

+

n

)

2

)

=

1

=

gcd

⁡

(

2

⁢

m

⁢

n

,

(

m

-

n

)

2

)

, and it follows that

gcd

⁡

(

2

⁢

m

⁢

n

,

m

+

n

)

=

1

,

gcd

⁡

(

2

⁢

m

⁢

n

,

m

-

n

)

=

1

. Thus

gcd

⁡

(

2

,

m

+

n

)

=

1

, i.e.

m

±

n

is odd. Furthermore,

gcd

⁡

(

2

⁢

m

⁢

n

,

m

+

n

)

=

1

implies

gcd

⁡

(

m

,

n

)

=

1

. And since the sum/

difference

of two

integers

is odd iff one is even, and the other is odd, only one of

m

,

n

is odd. Thus,

gcd

⁡

(

m

2

-

n

2

,

m

2

+

n

2

)

=

gcd

⁡

(

m

2

+

n

2

,

2

⁢

n

2

)

=

1

.

Conversely

, if

m

,

n

are coprime, and exactly one of

m

,

n

is odd, then

gcd

⁡

(

2

,

m

±

n

)

=

1

; thus,

2

⁢

m

⁢

n

,

m

2

-

n

2

are coprime. From the fact that

gcd

⁡

(

a

+

b

,

a

-

b

)

=

gcd

⁡

(

a

,

b

)

if

a

,

b

have opposite parity and

gcd

⁡

(

m

2

,

n

2

)

=

1

it follows that

m

2

-

n

2

,

m

2

+

n

2

are also coprime. And since

gcd

⁡

(

2

⁢

m

⁢

n

,

m

2

+

n

2

)

=

gcd

⁡

(

2

⁢

m

⁢

n

,

(

m

+

n

)

2

)

and

gcd

⁡

(

2

⁢

m

⁢

n

,

m

+

n

)

=

1

it follows that

2

⁢

m

⁢

n

,

m

2

+

n

2

are coprime. So the conditions the Pythagorean triple

{

2

⁢

m

⁢

n

,

m

2

-

n

2

,

m

2

+

n

2

}

is primitive,

gcd

⁡

(

2

⁢

m

⁢

n

,

m

+

n

)

=

1

and

m

,

n

are coprime and exactly one of them is odd are

equivalent

.

So if

a

,

b

,

c

satisfy

a

2

+

b

2

=

c

2

and

a

,

b

,

c

are pairwise coprime, then

c

is odd, and exactly one of

a

,

b

are odd and the other is even.

Let

n

,

m

be coprime

positive integers

of opposite parity,

n

<

m

. Set

n

′

=

m

+

n

,

m

′

=

n

-

m

(2)

in

equation

1

gives

b

=

m

′

⁢

n

′

,

c

=

n

2

+

m

2

2

=

m

′

2

+

n

′

2

2

,

a

=

n

′

2

-

m

′

2

2

(3)

since

n

=

m

′

+

n

′

2

,

m

=

n

′

-

m

′

2

. Clearly,

gcd

⁡

(

m

′

,

n

′

)

=

1

.

Now we prove that any

primitive Pythagorean triple

can be generated choosing

odd

coprime integers.

Remark 1

.

Let

m

,

n

be odd coprime integers,

n

<

m

. Let

f

1

=

n

2

-

m

2

2

,

f

2

=

n

2

+

m

2

2

. Then

gcd

⁡

(

f

1

,

f

2

)

=

1

.

Proof.

Since

gcd

⁡

(

n

2

,

m

2

)

=

1

=

gcd

⁡

(

f

1

+

f

2

,

f

1

-

f

2

)

, the statement follows from the fact that

f

1

,

f

2

have opposite parity since in this case

gcd

⁡

(

f

1

,

f

2

)

=

gcd

⁡

(

f

1

+

f

2

,

f

1

-

f

2

)

. Since

4

∣

n

2

-

m

2

,

f

2

is odd, and since

f

1

=

f

2

+

n

2

and

n

is odd,

f

1

,

f

2

have opposite parity.
∎

Substituting

C

=

n

2

,

B

=

m

2

in

B

⁢

C

=

(

C

+

B

2

)

2

-

(

C

-

B

2

)

2

yields that

m

⁢

n

,

n

2

-

m

2

2

,

n

2

+

m

2

2

is a primitive Pythagorean triple.

To see that any primitive Pythagorean triple is of this form:

Theorem 1

.

Let

a

,

b

be

positive

coprime integers satisfying

a

2

+

b

2

=

c

2

. Then

a

,

b

have opposite parity, and

c

is odd. Furthermore

(

a

,

c

)

=

(

b

,

c

)

=

1

.

Proof.

a

,

b

cannot both be even since

gcd

⁡

(

a

,

b

)

=

1

. If both

a

,

b

were odd we had

c

2

≡

2

(

mod

4

)

which is impossible since the square of any number is either

congruent

(

http://planetmath.org/

Congruences

) 0 or 1 modulo 4. Thus,

c

must be odd. Now for any integers

a

,

b

the congruence

a

2

+

b

2

≡

(

a

+

b

)

2

(

mod

2

)

holds. Together with

c

2

≡

1

(

mod

2

)

this gives

a

+

b

≡

1

(

mod

2

)

, so

a

,

b

have opposite parity.
∎

Suppose

a

is odd. Since

a

2

=

(

c

+

b

)

⁢

(

c

-

b

)

is a square, and

(

c

+

b

,

c

-

b

)

=

(

c

+

b

,

2

⁢

b

)

and

(

2

,

c

+

b

)

=

1

it follows that

c

±

b

are coprime and consequently each of them is square. This gives

c

-

b

=

n

2

,

c

+

b

=

m

2

where

m

,

n

are odd coprime integers, and we get

a

2

=

m

2

⁢

n

2

⇔

(4)

a

=

m

⁢

n

,

(5)

b

=

(

c

+

b

-

(

c

-

b

)

)

/

2

=

m

2

-

n

2

2

,

(6)

c

=

n

2

+

m

2

2

.

(7)

Now let

A

=

a

⁢

b

2

be a square. Without loss of generality we can set

a

=

m

⁢

n

,

b

=

n

2

-

m

2

2

where

m

,

n

are odd coprime integers. So we have

A

=

m

⁢

n

⁢

n

2

-

m

2

4

, and since

m

⁢

n

and

n

2

-

m

2

4

are coprime, each of them must itself be a square. So we have

c

2

=

(

n

2

+

m

2

2

)

2

=

(

m

⁢

n

)

2

+

(

n

2

-

m

2

2

)

2

(8)

where the right-hand side numbers are biquadratic integers. So the question if the area of a

right triangle

with integer sides is square is equivalent to asking if

x

4

+

y

4

=

z

2

has a solution in positive integers.

Title

proof of Pythagorean triplet

Canonical name

ProofOfPythagoreanTriplet

Date of creation

2013-03-22 14:06:52

Last modified on

2013-03-22 14:06:52

Owner

Thomas Heye (1234)

Last modified by

Thomas Heye (1234)

Numerical id

12

Author

Thomas Heye (1234)

Entry type

Proof

Classification

msc 11D09

Related topic

ContraharmonicMeansAndPythagoreanHypotenuses

Generated on Fri Feb 9 15:31:44 2018 by

LaTeXML