proof of Pythagorean triples

proof of Pythagorean triples

If

a

,

b

, and

c

are

positive integers

such that

a

2

+

b

2

=

c

2

(1)

then

(

a

,

b

,

c

)

is a

Pythagorean triple

. If

a

,

b

, and

c

are
relatively prime in pairs then

(

a

,

b

,

c

)

is a

primitive
Pythagorean triple

. Clearly, if

k

divides any two of

a

,

b

,
and

c

it divides all three. And if

a

2

+

b

2

=

c

2

then

k

2

⁢

a

2

+

k

2

⁢

b

2

=

k

2

⁢

c

2

. That is, for a positive integer

k

, if

(

a

,

b

,

c

)

is a Pythagorean triple then so is

(

k

⁢

a

,

k

⁢

b

,

k

⁢

c

)

.
Hence, to find all Pythagorean triples, it’s sufficient to find
all primitive Pythagorean triples.

Let

a

,

b

, and

c

be relatively prime positive integers such
that

a

2

+

b

2

=

c

2

. Set

m

n

=

a

+

c

b

reduced to

lowest terms

, That is,

gcd

⁡

(

m

,

n

)

=

1

. From the

triangle inequality

m

>

n

. Then

m

n

⁢

b

-

a

=

c

.

(2)

Squaring

both sides of (

2

) and multiplying through by

n

2

we get

m

2

⁢

b

2

-

2

⁢

m

⁢

n

⁢

a

⁢

b

+

n

2

⁢

a

2

=

n

2

⁢

a

2

+

n

2

⁢

b

2

;

which, after cancelling and rearranging terms, becomes

b

⁢

(

m

2

-

n

2

)

=

a

⁢

(

2

⁢

m

⁢

n

)

.

(3)

There are two cases, either

m

and

n

are of opposite parity, or
they or both odd. Since

gcd

⁡

(

m

,

n

)

=

1

, they can not both be
even.

Case 1.

m

and

n

of opposite parity, i.e.,

m

≢

±

n

⁢

(

m

⁢

o

⁢

d

⁢

2

)

. So 2 divides b since

m

2

-

n

2

is odd. From equation (

2

),

n

divides

b

.
Since

gcd

⁡

(

m

,

n

)

=

1

then

gcd

⁡

(

m

,

m

2

-

n

2

)

=

1

, therefore

m

also
divides

b

. And since

gcd

⁡

(

a

,

b

)

=

1

,

b

divides

2

⁢

m

⁢

n

. Therefore

b

=

2

⁢

m

⁢

n

. Then

a

=

m

2

-

n

2

,

b

=

2

⁢

m

⁢

n

,

and from
(

2

)

,

c

=

m

n

⁢

2

⁢

m

⁢

n

-

(

m

2

-

n

2

)

=

m

2

+

n

2

.

(4)

Case 2.

m

and

n

both odd, i.e.,

m

≡

±

n

⁢

(

m

⁢

o

⁢

d

⁢

2

)

. So 2 divides

m

2

-

n

2

. Then by the same process
as in the first case we have

a

=

m

2

-

n

2

2

,

b

=

m

⁢

n

,

a

⁢

n

⁢

d

c

=

m

2

+

n

2

2

.

(5)

The

parametric equations

in (

4

) and (

5

) appear
to be different but they generate the same solutions. To show
this, let

u

=

m

+

n

2

⁢

and

⁢

v

=

m

-

n

2

.

Then

m

=

u

+

v

, and

n

=

u

-

v

. Substituting those values for

m

and

n

into (

5

) we get

a

=

2

⁢

u

⁢

v

,

b

=

u

2

-

v

2

,

and

c

=

u

2

+

v

2

(6)

where

u

>

v

,

g

⁢

c

⁢

d

⁢

(

u

,

v

)

=

1

, and

u

and

v

are of opposite
parity. Therefore (

6

), with a and b
interchanged, is identical to (

4

). Thus since

(

m

2

-

n

2

,

2

⁢

m

⁢

n

,

m

2

+

n

2

)

, as in (

4

), is a
primitive Pythagorean triple, we can say that

(

a

,

b

,

c

)

is a
primitive pythagorean triple if and only if there exists
relatively prime, positive integers

m

and

n

,

m

>

n

, such that

a

=

m

2

-

n

2

,

b

=

2

⁢

m

⁢

n

,

and

⁢

c

=

m

2

+

n

2

.

Title

proof of Pythagorean triples

Canonical name

ProofOfPythagoreanTriples

Date of creation

2013-03-22 14:28:05

Last modified on

2013-03-22 14:28:05

Owner

fredlb (5992)

Last modified by

fredlb (5992)

Numerical id

9

Author

fredlb (5992)

Entry type

Proof

Classification

msc 11-00

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