x^4-y^4=z^2 has no solutions in positive integers

x

4

-

y

4

=

z

2

has no solutions in positive integers

We know (see example of Fermat’s Last

Theorem

) that the sum of two

fourth powers

can never be a square unless all are zero. This article shows that the

difference

of two fourth powers can never be a square unless at least one of the numbers is zero. Fermat proved this fact as part of his proof that the area of a

right triangle

with integral sides is never a square; see the corollary below. The proof of the main theorem is a great example of the method of

infinite descent

.

Theorem 1

.

x

4

-

y

4

=

z

2

has no solutions in

positive integers

.

Proof.

Suppose the equation has a solution in positive integers, and choose a solution that minimizes

x

2

+

y

2

. Note that

x

,

y

, and

z

are

pairwise coprime

, since otherwise we could divide out by their common

divisor

to get a smaller solution. Thus

z

2

+

(

y

2

)

2

=

(

x

2

)

2

so that

z

,

y

2

,

x

2

form a

pythagorean triple

. There are thus positive integers

p

,

q

of opposite parity (and

coprime

since

x

,

y

, and

z

are) such that

x

2

=

p

2

+

q

2

and either

y

2

=

p

2

-

q

2

or

y

2

=

2

⁢

p

⁢

q

.

Factoring

the original equation, we get

(

x

2

-

y

2

)

⁢

(

x

2

+

y

2

)

=

z

2

If

y

2

=

p

2

-

q

2

, then

(

x

⁢

y

)

2

=

p

4

-

q

4

, and clearly

p

2

+

q

2

=

x

2

<

x

2

+

y

2

, so we have found a solution smaller than the assumed minimal solution.

Assume therefore that

y

2

=

2

⁢

p

⁢

q

. Now,

x

2

=

p

2

+

q

2

; we may assume by relabeling if necessary that

q

is even and

p

odd. Then

p

,

q

,

x

are pairwise coprime and form a pythagorean triple; thus there are

P

>

Q

>

0

of opposite parity and coprime such that

q

=

2

⁢

P

⁢

Q

,

p

=

P

2

-

Q

2

,

x

=

P

2

+

Q

2

Then

P

⁢

Q

⁢

(

P

2

-

Q

2

)

=

1

2

⁢

p

⁢

q

=

y

2

4

is a square; it follows that

P

,

Q

, and

P

2

-

Q

2

are all (nonzero) squares since they are pairwise coprime. Write

P

=

R

2

,

Q

=

S

2

,

P

2

-

Q

2

=

T

2

for positive integers

R

,

S

,

T

. Then

T

2

=

R

4

-

S

4

, and

R

2

+

S

2

=

P

+

Q

<

(

P

+

Q

)

⁢

(

P

⁢

Q

)

⁢

(

P

-

Q

)

=

1

2

⁢

p

⁢

q

=

y

2

4

≤

y

2

<

x

2

+

y

2

We have thus found a smaller solution in positive integers, contradicting the

hypothesis

.
∎

Corollary 1

.

No right triangle with integral sides has area that is an integral square.

Proof.

Suppose

x

,

y

,

z

is a right triangle with

z

the

hypotenuse

, and let

d

=

gcd

⁡

(

x

,

y

,

z

)

. Either

x

/

d

or

y

/

d

is even; by relabeling if necessary, assume

x

/

d

is even. Then we can choose relatively

prime integers

p

,

q

with

p

>

q

and of opposite parity such that

x

=

(

2

⁢

p

⁢

q

)

⁢

d

y

=

(

p

2

-

q

2

)

⁢

d

z

=

(

p

2

+

q

2

)

⁢

d

If the triangle’s area is to be a square, then

1

2

⁢

x

⁢

y

=

p

⁢

q

⁢

(

p

2

-

q

2

)

⁢

d

2

must be a square, and thus

p

⁢

q

⁢

(

p

2

-

q

2

)

must be a square. Since

p

and

q

are coprime, it follows that

p

,

q

, and

p

2

-

q

2

are all squares, and thus that

p

2

-

q

2

is the difference of two fourth powers. But then

1

2

⁢

x

⁢

y

p

⁢

q

⁢

d

2

=

p

2

-

q

2

must also be a square. Since both

p

and

q

are squares, this is impossible by the theorem.
∎

Title

x

4

-

y

4

=

z

2

has no solutions in positive integers

Canonical

name

X4y4z2HasNoSolutionsInPositiveIntegers

Date of creation

2013-03-22 17:05:04

Last modified on

2013-03-22 17:05:04

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

7

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 11D41

Classification

msc 14H52

Classification

msc 11F80

Related topic

ExampleOfFermatsLastTheorem

Related topic

IncircleRadiusDeterminedByPythagoreanTriple

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