solutions of 1+x+x^2+x^3=y^2

solutions of

1

+

x

+

x

2

+

x

3

=

y

2

This article shows that the only solutions in

integers

to the equation

1

+

x

+

x

2

+

x

3

=

y

2

are the obvious trivial solutions

x

=

0

,

±

1

together with

x

=

7

,

y

=

20

. This result was known to Fermat.

First, note that the equation is

(

1

+

x

)

⁢

(

1

+

x

2

)

=

y

2

, so we have immediately that

x

≥

-

1

. So, noting the solutions for

x

=

0

,

±

1

, assume in what follows that

x

>

1

.

Let

d

=

gcd

⁡

(

1

+

x

,

1

+

x

2

)

. Then

x

≡

-

1

(

mod

d

)

, so that

1

+

x

2

≡

2

(

mod

d

)

. But

d

∣

1

+

x

2

so that

2

≡

0

(

mod

d

)

and

d

is either

1

or

2

. If

d

=

1

, so that

1

+

x

and

1

+

x

2

are

coprime

, then

1

+

x

and

1

+

x

2

must both be squares. But

1

+

x

2

is not a square for

x

>

1

. Thus the

gcd

of

1

+

x

and

1

+

x

2

is

2

, so each must be twice a square, say

1

+

x

=

2

⁢

r

2

1

+

x

2

=

2

⁢

s

2

and then

(

r

2

)

2

+

(

r

2

-

1

)

2

=

(

1

+

x

2

)

2

+

(

-

1

+

x

2

)

2

=

1

+

x

2

2

=

s

2

so that

r

2

,

r

2

-

1

,

s

form a

primitive Pythagorean triple

(note that

r

>

1

since

x

>

1

).

Recall that if

(

a

,

b

,

c

)

is a primitive Pythagorean triple, then precisely one of

a

and

b

is even, and we can choose coprime integers

p

,

q

such that

a

=

p

2

-

q

2

,

b

=

2

⁢

p

⁢

q

,

c

=

p

2

+

q

2

or

a

=

2

⁢

p

⁢

q

,

b

=

p

2

-

q

2

,

c

=

p

2

+

q

2

depending on the parity of

a

.

Suppose first that

r

is odd. Then

r

2

=

p

2

-

q

2

r

2

-

1

=

2

⁢

p

⁢

q

s

=

p

2

+

q

2

Then

1

=

r

2

-

(

r

2

-

1

)

=

(

p

-

q

)

2

-

2

⁢

q

2

. Now, note that

p

-

q

must be a square, say

p

-

q

=

t

2

, since

gcd

⁡

(

p

-

q

,

p

+

q

)

=

1

and

(

p

-

q

)

⁢

(

p

+

q

)

is a square. Then

t

4

=

(

p

-

q

)

2

=

2

⁢

q

2

+

1

=

(

q

2

+

1

)

2

-

q

4

so that

t

4

+

q

4

=

(

q

2

+

1

)

2

But we know that the sum of two

fourth powers

can be a square (

http://

planetmath

.org/ExampleOfFermatsLastTheorem

) only for the trivial case where all are zero. So

r

cannot be odd.

So suppose that

r

is even. Then

r

2

=

2

⁢

p

⁢

q

r

2

-

1

=

p

2

-

q

2

s

=

p

2

+

q

2

From the second of these formulas, we see that

p

must be even (consider both sides

(

mod

4

)

), say

p

=

2

⁢

t

2

. Now,

(

p

+

q

-

1

)

⁢

(

p

+

q

+

1

)

=

(

p

+

q

)

2

-

1

=

p

2

+

2

⁢

p

⁢

q

+

q

2

-

1

=

p

2

+

2

⁢

p

⁢

q

+

q

2

-

(

r

2

-

(

r

2

-

1

)

)

=

p

2

+

2

⁢

p

⁢

q

+

q

2

-

2

⁢

p

⁢

q

+

p

2

-

q

2

=

2

⁢

p

2

=

8

⁢

t

4

Since

p

and

q

have opposite parity,

p

+

q

±

1

are even, so that

2

⁢

t

4

=

p

+

q

-

1

2

⋅

(

p

+

q

-

1

2

+

1

)

=

u

⁢

(

u

+

1

)

Thus one of

u

,

u

+

1

is a fourth power and the other is twice a fourth power, say

b

4

and

2

⁢

c

4

.

Now,

u

=

b

4

⇒

u

+

1

=

2

⁢

c

4

⇒

b

4

-

2

⁢

c

4

=

-

1

u

=

2

⁢

c

4

⇒

u

+

1

=

b

4

⇒

b

4

-

2

⁢

c

4

=

1

so that

b

4

-

2

⁢

c

4

=

±

1

.

If

b

4

-

2

⁢

c

4

=

1

, then

(

(

c

2

)

2

+

1

)

2

=

(

c

2

)

4

+

b

4

and again we have a square being the sum of two fourth powers. So this case is impossible.

If

b

4

-

2

⁢

c

4

=

-

1

, write

e

=

c

2

, then

(

e

2

-

1

)

2

=

e

4

-

b

4

. It follows (see here (

http://planetmath.org/X4Y4z2HasNoSolutionsInPositiveIntegers

)) that either

b

=

0

or

e

2

-

1

=

0

. If

b

=

0

, we get the impossibility

e

4

-

2

⁢

e

2

+

1

=

e

4

, while if

(

e

2

-

1

)

2

=

0

, then

e

4

=

b

4

,

e

=

±

1

and so

b

=

±

1

. Then

p

+

q

-

1

2

=

b

4

=

1

, so that

p

+

q

=

3

. Thus

r

2

=

4

, so

r

=

2

and, finally,

1

+

x

=

2

⁢

r

2

=

8

, so that

x

=

7

.

Thus, the only nontrivial solution to the equation given is

1

+

7

+

49

+

343

=

400

Title

solutions of

1

+

x

+

x

2

+

x

3

=

y

2

Canonical name

SolutionsOf1xx2x3y2

Date of creation

2013-03-22 17:05:09

Last modified on

2013-03-22 17:05:09

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

10

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 11F80

Classification

msc 14H52

Classification

msc 11D41

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