infinite descent

infinite descent

Fermat invented this method of

infinite descent

. The idea is: If a given

natural number

n

with certain properties implies that there exists a smaller one with these properties, then
there are infinitely many of these, which is impossible.

Here is an example:

Let

m

,

n

be

coprime

positive integers

with opposite parity,

m

<

n

, and, say,

m

is even.

Let

a

=

2

⁢

m

⁢

n

,

b

=

n

2

-

m

2

,

c

=

m

2

+

n

2

. Then

{

a

,

b

,

c

}

is a

primitive Pythagorean triple

,
and the area

A

of the

right triangle

with sides

a

,

b

,

c

is

a

⁢

b

/

2

=

m

⁢

n

⁢

(

n

2

-

m

2

)

.

Suppose

A

is a square. Then, since

m

,

n

are coprime and of opposite parity,

gcd

⁡

(

m

+

n

,

m

-

n

)

=

gcd

⁡

(

m

,

n

)

=

1

. Thus, for

A

to be a square, each of

m

,

n

,

m

-

n

,

m

+

n

must be
squares itself. Setting

r

2

=

m

,

s

2

=

n

, we have

A

=

(

r

⁢

s

)

2

⁢

(

s

4

-

r

4

)

.

We prove that the

Diophantine equation

x

4

-

y

4

=

z

2

has no solution in natural numbers.

Remark 1

.

Suppose that

z

2

+

y

4

=

x

4

, where

gcd

⁡

(

x

,

y

,

z

)

=

1

,

x

,

y

,

z

∈

N

. Then

x

is odd, and

y

,

z

have opposite parity.

Proof.

If

x

was even, then

x

4

=

z

2

+

y

4

≡

(

z

+

y

2

)

2

≡

0

(

mod

2

)

, so

z

,

y

2

≡

0

(

mod

2

)

or

z

,

y

2

≡

1

(

mod

2

)

. But

z

,

y

2

≡

0

(

mod

2

)

conflicts with

gcd

⁡

(

x

,

y

,

z

)

=

1

. And

z

,

y

2

≡

1

(

mod

2

)

implies

y

2

+

(

z

2

)

2

≡

2

(

mod

4

)

contradicting

x

4

≡

0

(

mod

4

)

. Thus,

x

is odd, and

x

4

=

z

2

+

(

y

2

)

2

≡

(

z

+

y

2

)

2

≡

1

(

mod

2

)

implies that

z

,

y

2

have opposite parity.
∎

Suppose

x

is odd and

z

is even. Then we have

z

=

2

⁢

p

⁢

q

,

y

2

=

q

2

-

p

2

and

x

2

=

q

2

+

p

2

, where

p

,

q

have opposite parity and are coprime. Since

z

is odd, this implies

(

x

⁢

y

)

2

=

q

4

-

p

4

, so it is

sufficient

to show that there is no solution for odd

z

.

Now

x

,

z

are assumed odd. Then

y

is even, and there exist

m

,

n

∈

ℕ

,

m

<

n

,

(

2

⁢

m

⁢

n

,

m

+

n

)

=

1

such that

y

2

=

2

⁢

m

⁢

n

(1)

x

2

=

n

2

+

m

2

(2)

z

=

n

2

-

m

2

.

(3)

Since

m

2

+

n

2

=

x

2

is a primitive Pythagorean triple, there exist

p

,

q

∈

ℕ

,

p

<

q

,

(

2

⁢

p

⁢

q

,

p

+

q

)

=

1

satisfiying

m

=

2

⁢

p

⁢

q

(4)

n

=

q

2

-

p

2

(5)

x

=

q

2

+

p

2

.

(6)

Since

2

⁢

m

⁢

n

is a square and

m

,

n

are coprime and, say,

n

is odd,

n

is a square, and we have

m

=

2

⁢

r

2

,

n

=

s

2

.

From the primitive Pythagorean triple

m

2

+

n

2

=

x

2

we get

x

=

u

2

+

v

2

,

n

=

u

2

-

v

2

,

m

=

2

⁢

u

⁢

v

. Since

2

⁢

u

⁢

v

=

2

⁢

r

2

u

⁢

v

is a square, and each of

u

and

v

is a square:

u

=

g

2

,

v

=

h

2

.

Substituting

n

,

u

,

v

in

n

=

u

2

-

v

2

we have

s

2

=

g

4

-

h

4

.
But since

n

+

(

1

/

2

)

<

m

this implies

n

=

s

2

<

z

=

n

2

-

m

2

<

z

2

, thus we have another solution with odd

s

<

z

. This contradicts to the fact that there exists a smallest solution.

See

http://mathpages.com/home/kmath144.htm

here for a discussion of
infinite descent vs.

induction

.

Title

infinite descent

Canonical name

InfiniteDescent

Date of creation

2013-03-22 14:07:56

Last modified on

2013-03-22 14:07:56

Owner

Thomas Heye (1234)

Last modified by

Thomas Heye (1234)

Numerical id

13

Author

Thomas Heye (1234)

Entry type

Topic

Classification

msc 11D25

Related topic

ExampleOfFermatsLastTheorem

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