EXAMPLES OF MORDELL’S EQUATION
KEITH CONRAD
1. Introduction
The equation y
2
= x
3
+ k, for k ∈ Z, is called Mordell’s equation
1
due to Mordell’s work
on it throughout his life. A natural number-theoretic task is describing all of its solutions
in Z or Q, either qualitatively (decide if there are finitely or infinitely many solutions in Z
or Q) or quantitatively (list or otherwise conveniently describe all such solutions). In 1920,
Mordell [10] showed that for each nonzero k ∈ Z, y
2
= x
3
+ k has finitely many integral
solutions. Rational solutions are a different story: there may be finitely or infinitely many,
depending on k. Whether there are finitely or infinitely many rational solutions is connected
to a central topic in number theory: the rank of an elliptic curve.
Here we will describe all integral solutions to Mordell’s equation for some selected values
of k,
2
and make a few comments at the end about rational solutions. For further examples
of the techniques we use to find integral solutions, see [1, Chap. 14].
2. Examples without Solutions
To prove y
2
= x
3
+k has no integral solution for specific values of k, we will use congruence
and quadratic residue considerations. Specifically, we will use the following descriptions of
when −1, 2, and −2 are squares modulo odd primes p, writing “” to mean a square:
−1 ≡ mod p ⇐⇒ p ≡ 1 mod 4,
2 ≡ mod p ⇐⇒ p ≡ 1, 7 mod 8,
−2 ≡ mod p ⇐⇒ p ≡ 1, 3 mod 8.
Our first three theorems will use the criterion for −1 ≡ mod p.
Theorem 2.1. The equation y
2
= x
3
+ 7 has no integral solutions.
Proof. Assume there is an integral solution (x, y). If x is even then y
2
≡ 7 mod 8, but
7 mod 8 is not a square. Therefore x is odd. Rewrite y
2
= x
3
+ 7 as y
2
+ 1 = x
3
+ 8, so
(1) y
2
+ 1 = x
3
+ 8 = (x + 2)(x
2
− 2x + 4).
Note x
2
−2x + 4 = (x −1)
2
+ 3 is at least 3. Since x is odd, (x − 1)
2
+ 3 ≡ 3 mod 4. Thus
x
2
− 2x + 4 has a prime factor p ≡ 3 mod 4: if not, all of its prime factors are 1 mod 4,
so x
2
− 2x + 4 ≡ 1 mod 4 since a positive integer is the product of its prime factors (this
isn’t true for −5: −5 ≡ 3 mod 4 and the prime factor of −5 is 1 mod 4). That contradicts
x
2
−2x+4 ≡ 3 mod 4. From p | (x
2
−2x+4) we get p | (y
2
+1) by (1), so y
2
+1 ≡ 0 mod p.
Thus −1 ≡ mod p, contradicting p ≡ 3 mod 4. This is V. A. Lebesgue’s method [8].
1
Also called Bachet’s equation.
2
Large tables of k and their integral solutions are at https://hr.userweb.mwn.de/numb/mordell.html
and https://secure.math.ubc.ca/∼bennett/BeGa-data.html.
1
2 KEITH CONRAD
Here’s another approach, using the factor x + 2 instead of the factor x
2
− 2x + 4. Since
(as seen above) x is odd and y is even, x
3
≡ x mod 4 (true for all odd x), so reducing
y
2
= x
3
+ 7 modulo 4 gives us 0 ≡ x + 3 mod 4, so x ≡ 1 mod 4. Then x + 2 ≡ 3 mod 4.
Moreover, x + 2 > 0, since if x ≤ −2 then x
3
≤ −8, so x
3
+7 ≤ −1, which contradicts x
3
+7
being a perfect square. From x + 2 being positive and congruent to 3 mod 4, it has a prime
factor p ≡ 3 mod 4, so y
2
+ 1 ≡ 0 mod p from (1) and we get a contradiction as before.
Theorem 2.2. The equation y
2
= x
3
− 5 has no integral solutions.
Proof. Assuming there is a solution, reduce modulo 4:
y
2
≡ x
3
− 1 mod 4.
Here is a table of values of y
2
and x
3
− 1 modulo 4:
y y
2
mod 4 x x
3
− 1 mod 4
0 0 0 3
1 1 1 0
2 0 2 3
3 1 3 2
The only common value of y
2
mod 4 and x
3
−1 mod 4 is 0, so y is even and x ≡ 1 mod 4.
Then rewrite y
2
= x
3
− 5 as
(2) y
2
+ 4 = x
3
− 1 = (x − 1)(x
2
+ x + 1).
Since x ≡ 1 mod 4, x
2
+ x + 1 ≡ 3 mod 4, so x
2
+ x + 1 is odd. Moreover, x
2
+ x + 1 =
(x + 1/2)
2
+ 3/4 > 0, so x
2
+ x + 1 ≥ 3. Therefore x
2
+ x + 1 must have a prime factor
p ≡ 3 mod 4 (same reasoning as in the previous proof). Since p is a factor of x
2
+ x + 1, p
divides y
2
+ 4 by (2), so y
2
+ 4 ≡ 0 mod p. Therefore −4 ≡ mod p, so −1 ≡ mod p.
This implies p ≡ 1 mod 4, contradicting p ≡ 3 mod 4.
Theorem 2.3. The equation y
2
= x
3
+ 11 has no integral solutions.
Proof. We will use ideas from the proofs of Theorems 2.1 and 2.2.
Assume there is an integral solution (x, y). Since 11 ≡ −1 mod 4, the same reasoning as
in the proof of Theorem 2.2 shows x ≡ 1 mod 4.
Rewrite y
2
= x
3
+ 11 as
(3) y
2
+ 16 = x
3
+ 27 = (x + 3)(x
2
− 3x + 9).
The factor x
2
−3x+9 is positive (why?), and from x ≡ 1 mod 4 we get x
2
−3x+9 ≡ 3 mod 4,
so by the same reasoning as in the proof of Theorem 2.1, x
2
− 3x + 9 has a prime factor
p with p ≡ 3 mod 4. Therefore p | (y
2
+ 16) by (3), so −16 ≡ mod p. Since p is odd,
−1 ≡ mod p, and that contradicts p ≡ 3 mod 4.
Our next two theorems will rely on the condition for when 2 ≡ mod p.
Theorem 2.4. The equation y
2
= x
3
− 6 has no integral solutions.
Proof. Assume there is an integral solution. If x is even then y
2
≡ −6 ≡ 2 mod 8, but
2 mod 8 is not a square. Therefore x is odd, so y is odd and x
3
= y
2
+ 6 ≡ 7 mod 8. Also
x
3
≡ x mod 8 (true for all odd x), so x ≡ 7 mod 8.
Rewrite y
2
= x
3
− 6 as
(4) y
2
− 2 = x
3
− 8 = (x − 2)(x
2
+ 2x + 4),
EXAMPLES OF MORDELL’S EQUATION 3
with x
2
+ 2x + 4 ≡ 7
2
+ 2 · 7 + 4 ≡ 3 mod 8. Since x
2
+ 2x + 4 = (x + 1)
2
+ 3 is positive,
it must have a prime factor p ≡ ±3 mod 8 because if all of its prime factors are ±1 mod 8
then x
2
+ 2x + 4 ≡ ±1 mod 8, which is not true. Let p be a prime factor of x
2
+ 2x + 4 with
p ≡ ±3 mod 8. Since p divides y
2
− 2 by (4), we get y
2
≡ 2 mod p. Thus 2 ≡ mod p, so
p ≡ ±1 mod 8, which is a contradiction.
We can get a contradiction using the factor x−2 also. Since x ≡ 7 mod 8, x−2 ≡ 5 mod 8.
Also x −2 > 0, since if x ≤ 2 and x −2 ≡ 5 mod 8 then x ≤ −1, but then x
3
−6 is negative
so it can’t be a perfect square. From x −2 being positive and congruent to 5 mod 8, it has a
prime factor p ≡ ±3 mod 8 and then y
2
≡ 2 mod p and we get a contradiction in the same
way as before.
Theorem 2.5. The equation y
2
= x
3
+ 45 has no integral solutions.
Proof. Assume there is an integral solution. If y is odd then x
3
= y
2
−45 ≡ 1−45 ≡ 4 mod 8,
which is impossible. Therefore y is even, so x is odd. Reducing the equation mod 4,
0 ≡ x
3
+ 1 mod 4. Since x
3
≡ x mod 4 for odd x, x ≡ 3 mod 4. Also, y is not a multiple of
3. If 3 | y then the equation y
2
= x
3
+ 45 shows 3 divides x. Write x = 3x
0
and y = 3y
0
, so
9y
02
= 27x
03
+ 45, so y
02
= 3x
03
+ 5, which implies y
02
≡ 2 mod 3, and that is impossible.
We will now take cases depending on whether x ≡ 3 mod 8 or x ≡ 7 mod 8. (If you know
an elementary method that treats both cases in a uniform way, please tell me!)
Case 1: x ≡ 3 mod 8. Rewrite y
2
= x
3
+ 45 as
(5) y
2
− 72 = x
3
− 27 = (x − 3)(x
2
+ 3x + 9).
The factor x
2
+ 3x + 9 = (x + 3/2)
2
+ 27/4 is positive and is congruent to 3 mod 8, so it
has a prime factor p ≡ ±3 mod 8. Feeding this into (5),
(6) y
2
≡ 72 ≡ 2 · 6
2
mod p.
We can’t have p = 3 (just in case p ≡ 3 mod 8, this is something we need to deal with)
since it would imply y
2
≡ 0 mod 3, but we already checked y is not a multiple of 3. Since
p is not 3, (6) implies 2 ≡ mod p, so p ≡ ±1 mod 8, contradicting p ≡ ±3 mod 8.
Case 2: x ≡ 7 mod 8. Rewrite y
2
= x
3
+ 45 as
(7) y
2
− 18 = x
3
+ 27 = (x + 3)(x
2
− 3x + 9).
The factor x
2
− 3x + 9 = (x − 3/2)
2
+ 27/4 is positive and is congruent to 5 mod 8, so it
has a prime factor p ≡ ±3 mod 8. From (7) we get y
2
≡ 18 ≡ 2 · 3
2
mod p. Arguing as in
Case 1, we again find p ≡ ±1 mod 8, which is a contradiction.
In our next two theorems we will use the condition for when −2 ≡ mod p.
Theorem 2.6. The equation y
2
= x
3
+ 6 has no integral solutions.
Proof. Mordell [11, p. 22-23], [12, p. 70] proved this using Z[
√
6]. The simpler method used
here, which resembles the proof of Theorem 2.4, is due to Shiv Gupta and Tracy Driehaus.
Assume there is an integral solution. First we will show x is odd, and in fact x ≡ 3 mod 8.
If x is even then y
2
≡ 6 mod 8, which is impossible. Thus x is odd, so y is odd and
x
3
= y
2
− 6 ≡ −5 ≡ 3 mod 8. Since x
3
≡ x mod 8, we have x ≡ 3 mod 8.
Rewrite y
2
= x
3
+ 6 as
(8) y
2
+ 2 = x
3
+ 8 = (x + 2)(x
2
− 2x + 4),
with x
2
− 2x + 4 ≡ 3
2
− 2 · 3 + 4 ≡ 7 mod 8. For each prime factor p of x
2
− 2x + 4,
y
2
+2 ≡ 0 mod p, so −2 ≡ mod p, and therefore p ≡ 1, 3 mod 8. Then since x
2
−2x + 4 =
4 KEITH CONRAD
(x − 1)
2
+ 3 is positive, x
2
−2x + 4 is 1 or 3 mod 8. We showed before that this number is
7 mod 8, so we have a contradiction.
To get a contradiction using the factor x + 2, first note that this number is positive,
since if x + 2 < 0 then y
2
+ 2 ≤ 0, which is impossible. For a prime p dividing x + 2,
y
2
+ 2 ≡ 0 mod p, so p ≡ 1 or 3 mod 8. Therefore x + 2 ≡ 1 or 3 mod 8. However, since
x ≡ 3 mod 8 we have x + 2 ≡ 5 mod 8, which is a contradiction.
Theorem 2.7. The equation y
2
= x
3
+ 46 has no integral solutions.
Proof. Assume there is an integral solution. If x is even then y
2
≡ 6 mod 8, which has no
solution, so x is odd and y is odd. Thus y
2
≡ 1 mod 8 and x
3
≡ x mod 8, so 1 ≡ x+6 mod 8,
making x ≡ 3 mod 8.
Now rewrite y
2
= x
3
+ 46 as
(9) y
2
+ 18 = x
3
+ 64 = (x + 4)(x
2
− 4x + 16).
Since x ≡ 3 mod 8, the first factor on the right side of (9) is 7 mod 8 and the second factor
is 5 mod 8. We will get a contradiction using either of these factors.
First we work with the quadratic factor x
2
− 4x + 16 = (x − 2)
2
+ 12, which is positive.
Since it is 5 mod 8, it must have a prime factor p that is not 1 or 3 mod 8. Indeed, if all the
prime factors of x
2
−4x + 16 are 1 or 3 mod 8 then so is x
2
−4x + 16, since {1, 3 mod 8} is
closed under multiplication. But x
2
−4x+16 6≡ 1, 3 mod 8. The prime p, not being 3 mod 8,
is in particular not equal to 3. Also, p 6= 2 since x
2
−4x+16 is odd. Since p | (x
2
−4x+16) we
get by (9) that p | (y
2
+ 18), so y
2
≡ −18 mod p. Hence −18 ≡ mod p, so −2 ≡ mod p.
This implies p ≡ 1 or 3 mod 8. But p 6≡ 1 or 3 mod 8, so we have a contradiction.
To get a contradiction using the factor x + 4, first let’s check it is positive. There
is no solution to y
2
= x
3
+ 46 when y
2
is a perfect square less than 46 (just try y
2
=
0, 1, 4, 9, 16, 25, 36; there is no corresponding integral x), which means we must have x
3
> 0,
so x > 0. Thus x + 4 > 1. Since x + 4 ≡ 7 mod 8, x + 4 must have a prime factor p that is
not 1 or 3 mod 8, just as before. The prime p is not 2 since x + 4 is odd, and p 6= 3 since
p 6≡ 3 mod 8. Then y
2
≡ −18 mod p from (9) and we get a contradiction as before.
Our next theorems uses the conditions for −1 mod p and −2 mod p to be squares.
Theorem 2.8. The equation y
2
= x
3
− 24 has no integral solutions.
Proof. We take our argument from [13, pp. 271–272], which is based on [14, p. 201].
Assuming there is an integral solution (x, y), we show x is even. Rewrite y
2
= x
3
−24 as
y
2
+ 16 = x
3
− 8 = (x − 2)(x
2
+ 2x + 4).
The factor x
2
+ 2x + 4 equals (x + 1)
2
+ 3, which is at least 3. If x is odd then (x + 1)
2
+ 3 ≡
3 mod 4, so (x + 1)
2
+ 3 has a prime factor p such that p ≡ 3 mod 4. Then y
2
≡ −16 mod p,
so −1 ≡ mod p. This contradicts the condition p ≡ 3 mod 4. Therefore x is even, so also
y is even.
From y
2
= x
3
− 24 we get 8 | y
2
, so 4 | y. Write x = 2x
0
and y = 4y
0
. Then 16y
02
=
8x
03
−24, which implies 2y
02
= x
03
−3, so x
0
is odd and greater than 1. Rewrite 2y
02
= x
03
−3
as
2(y
02
+ 2) = x
03
+ 1 = (x
0
+ 1)(x
02
− x
0
+ 1).
The factor x
02
− x
0
+ 1 is odd and greater than 1. Let p be a prime factor of it, so y
02
≡
−2 mod p, which implies p ≡ 1 or 3 mod 8. Then x
02
−x
0
+ 1 is a product of primes that are
all 1 or 3 mod 8, so x
02
−x
0
+1 ≡ 1 or 3 mod 8. But y
02
≡ 0, 1, or 4 mod 8 ⇒ x
03
= 2y
02
+3 ≡
EXAMPLES OF MORDELL’S EQUATION 5
3 or 5 mod 8 ⇒ x
0
≡ 3 or 5 mod 8. Then x
02
−x
0
+ 1 ≡ 1 − x
0
+ 1 ≡ 2 − x
0
≡ 5 or 7 mod 8.
That contradicts x
02
− x
0
+ 1 ≡ 1 or 3 mod 8.
As an exercise, show each of the following has no integral solutions by methods like those
used above. In each case, begin by showing x is odd (this is trickier for the third equation).
(1) y
2
= x
3
− 3 (Hint: y
2
+ 4 = x
3
+ 1).
(2) y
2
= x
3
− 9 (Hint: y
2
+ 1 = x
3
− 8).
(3) y
2
= x
3
− 12 (Hint: y
2
+ 4 = x
3
− 8).
3. Examples with Solutions
We will now look at some instances of Mordell’s equation that have integral solutions.
The goal in each case is to find all the integral solutions. The main tool we will use is
unique factorization (in different settings), and after some successes we will see that this
technique eventually runs into difficulties.
We start with the case k = 16: the equation y
2
= x
3
+ 16. There are two obvious integral
solutions: (x, y) = (0, ±4). A numerical search does not reveal additional integral solutions,
so one might guess
3
that (0, 4) and (0, −4) are the only integral solutions. To prove this,
we will use unique factorization in Z.
Theorem 3.1. The only integral solutions to y
2
= x
3
+ 16 are (x, y) = (0, ±4).
Proof. First we determine the parity of an integral solution. Rewrite the equation as x
3
=
y
2
−16 = (y + 4)(y −4). If y is odd then (y + 4, y −4) = 1 (why?), so both y + 4 and y −4
are cubes because their product is a cube. They differ by 8, and no odd cubes differ by 8.
Hence y is even, so x is even.
The right side of y
2
= x
3
+ 16 is divisible by 8, so 4 | y. Writing y = 4y
0
, 16y
02
= x
3
+ 16.
Therefore 4 | x. Write x = 4x
0
, so y
02
= 4x
03
+ 1, showing y
0
is odd. Write y
0
= 2m + 1, so
4m
2
+4m+1 = 4x
03
+1. Thus m
2
+m = x
03
. Since m
2
+m = m(m+1) and (m, m+1) = 1,
the product m(m + 1) being a cube implies (since ±1 are cubes) both m and m + 1 are
cubes. The only consecutive cubes are among {−1, 0, 1}, so m or m + 1 is 0. Therefore
x
0
= 0, so x = 0 and y = ±4.
For the next few results, we use unique factorization in Z[i] or Z[
√
−2].
Theorem 3.2. The only x, y ∈ Z satisfying y
2
= x
3
− 1 is (x, y) = (1, 0).
Proof. First we check the parity of an integral solution. Suppose x is even, so y
2
+ 1 = x
3
≡
0 mod 8. Then y
2
≡ −1 mod 8. But −1 mod 8 is not a square. We have a contradiction, so
x is odd, which means y has to be even.
Write the equation y
2
= x
3
− 1 as
x
3
= y
2
+ 1,
which in Z[i] has the factored form
(10) x
3
= (y + i)(y − i).
If the two factors on the right side are relatively prime in Z[i], then since their product
is a cube, each factor must be a cube up to unit multiple, by unique factorization in Z[i].
3
It’s a tricky business to decide when to stop searching: y
2
= x
3
+24 has integral solutions at x = −2, 1, 10,
and 8158 (and no others). Contrast that with y
2
= x
3
− 24 in Theorem 2.8. The equation y
2
= x
3
− 999
has integral solutions at x = 10, 12, 40, 147, 174, and 22480 (and no others) [17].
6 KEITH CONRAD
Moreover, since every unit in Z[i] is a cube (1 = 1
3
, −1 = (−1)
3
, i = (−i)
3
, −i = i
3
), unit
factors can be absorbed into the cubes. Thus, provided we show y +i and y −i are relatively
prime, (10) tells us y + i and y − i are themselves cubes.
To see that y +i and y −i are relatively prime, let δ be a common divisor. Since δ divides
(y + i) − (y − i) = 2i, N(δ) divides N(2i) = 4. Also N(δ) divides N(y + i) = y
2
+ 1 = x
3
,
which is odd. Therefore N(δ) divides 4 and is odd, which means N(δ) = 1, so δ is a unit.
Now that we know y + i and y −i are relatively prime, we must have (as argued already)
y + i = (m + ni)
3
for some m, n ∈ Z. Expanding the cube and equating real and imaginary parts,
y = m
3
− 3mn
2
= m(m
2
− 3n
2
), 1 = 3m
2
n − n
3
= n(3m
2
− n
2
).
The equation on the right tells us n = ±1. If n = 1, then 1 = 3m
2
− 1, so 3m
2
= 2, which
has no integer solution. If n = −1, then 1 = −(3m
2
− 1), so m = 0. Therefore y = 0, so
x
3
= y
2
+ 1 = 1. Thus x = 1.
Theorem 3.3. The only x, y ∈ Z satisfying y
2
= x
3
− 4 are (x, y) = (2, ±2) and (5, ±11).
Proof. We rewrite y
2
= x
3
− 4 in Z[i] as
(11) x
3
= y
2
+ 4 = (y + 2i)(y − 2i).
We will show that both factors on the right are cubes. Let’s first see why this leads to the
desired integral solutions. Write
y + 2i = (m + ni)
3
for some m, n ∈ Z. Equating real and imaginary parts,
y = m(m
2
− 3n
2
), 2 = n(3m
2
− n
2
).
From the second equation, n = ±1 or n = ±2. In each case we try to solve for m in Z. The
cases that work out are n = 1 and m = ±1, and n = −2 and m = ±1. In the first case,
y = ±(1 − 3) = ±2 and x = 2, while in the second case y = ±(1 − 3 · 4) = ±11 and x = 5.
It remains to show in (11) that y + 2i and y − 2i are cubes. Since y
2
≡ x
3
mod 2 either
x and y are both even or they are both odd. We will consider these cases separately, since
they affect the greatest common factor of y + 2i and y − 2i.
First suppose x and y are both odd. We will show y + 2i and y − 2i are relatively prime
in Z[i]. Let δ be a common divisor, so δ divides (y + 2i) − (y − 2i) = 4i. Therefore N(δ)
divides N(4i) = 16. Since N(δ) also divides N(y + 2i) = y
2
+ 4 = x
3
, which is odd, we must
have N(δ) = 1, so δ is a unit. This means y + 2i and y − 2i are relatively prime, so since
their product in (11) is a cube and every unit in Z[i] is a cube, y + 2i and y − 2i are both
cubes.
Now suppose x and y are both even. Write x = 2x
0
and y = 2y
0
, so 4y
02
= 8x
03
− 4.
Dividing by 4, y
02
= 2x
03
− 1. Therefore y
0
is odd. We must have x
0
odd too, as otherwise
y
02
≡ −1 mod 4, but −1 mod 4 is not a square. Writing
2x
03
= y
02
+ 1 = (y
0
+ i)(y
0
− i),
the factors on the right each have even norm, so each is divisible by 1 + i. Divide the
equation by (1 + i)
2
= 2i:
−ix
03
=
y
0
+ i
1 + i
y
0
− i
1 + i
.
EXAMPLES OF MORDELL’S EQUATION 7
We will show the two factors on the right are relatively prime. Their difference is 2i/(1+i) =
1 + i, so each common divisor has norm dividing N(1 + i) = 2. Also each common divisor
divides x
03
, so the norm divides N(x
03
) = x
06
, which is odd. Thus each common divisor
of (y
0
+ i)/(1 + i) and (y
0
− i)/(1 + i) has norm 1, so is a unit. As before, we now know
(y
0
+ i)/(1 + i) is a cube, so
y + 2i = 2(y
0
+ i) = −i(1 + i)
2
(y
0
+ i) = i
3
(1 + i)
3
y
0
+ i
1 + i
is a cube in Z[i]. Similarly, y − 2i is a cube.
Theorem 3.4. The only integral solutions to y
2
= x
3
− 2 are (x, y) = (3, ±5).
Proof. Suppose y
2
= x
3
− 2 with integral x and y. As in the previous proof, first we do a
parity check on x and y. If x is even then y
2
≡ −2 mod 8, but −2 mod 8 is not a square.
Therefore x is odd, so y is also odd.
Write the relation between x and y as
x
3
= y
2
+ 2.
In Z[
√
−2], we can rewrite this as
(12) x
3
= (y +
√
−2)(y −
√
−2).
The two factors on the right are relatively prime. Indeed, let δ be a common divisor,
so δ divides their difference (y +
√
−2) − (y −
√
−2) = 2
√
−2, which means N(δ) divides
N(2
√
−2) = 8. At the same time, N(δ) divides N(y +
√
−2) = y
2
+ 2, which is odd since y is
odd. So N(δ) must be 1, which means δ is a unit in Z[
√
−2], so y +
√
−2 and y −
√
−2 are
relatively prime in Z[
√
−2]. From (12) and unique factorization in Z[
√
−2], y +
√
−2 and
y −
√
−2 are both cubes up to unit multiple. The units in Z[
√
−2] are ±1, which are both
cubes, and therefore a unit multiple of a cube is also a cube. Hence y +
√
−2 and y −
√
−2
are both cubes.
Write
y +
√
−2 = (m + n
√
−2)
3
for some m, n ∈ Z. It follows that
y = m
3
− 6mn
2
= m(m
2
− 6n
2
), 1 = 3m
2
n − 2n
3
= n(3m
2
− 2n
2
).
From the second equation, n = ±1. When n = 1 the second equation says 1 = 3m
2
− 2,
so m = ±1. Then y = ±1(1 − 6) = ±5 and x
3
= y
2
+ 2 = 27, so we recover the solutions
(x, y) = (3, ±5). When n = −1 we have 1 = −(3m
2
− 2 · 1
2
) = −(3m
2
− 2), so 1 = 3m
2
,
which has no solution in Z.
Remark 3.5. Theorems 3.3 and 3.4 were challenges by Fermat to British mathematicians
[4, p. 533], [18, pp. 103, 113]. Fermat said he could solve them by infinite descent, but gave
no details. Our proof of Theorems 3.3 and 3.4, like that of Theorem 3.2, studies integers of
the form y
2
+ n by factoring them in Z[
√
−n], which is an idea due to Euler.
Our treatment of y
2
= x
3
+ 16, y
2
= x
3
− 1, y
2
= x
3
− 4, and y
2
= x
3
− 2 relied on
features of Z, Z[i], and Z[
√
−2]: they have unique factorization and each unit in them is a
cube.
As an exercise, show each of the following three Mordell equations has only the indicated
integral solutions by using methods like those above.
(1) y
2
= x
3
− 8 in Z only for (x, y) = (2, 0). (Hint: Start by showing y is even.)
8 KEITH CONRAD
(2) y
2
= x
3
− 16 has no integral solutions. (Hint: Start by showing y is odd.)
(3) y
2
= x
3
− 64 in Z only for (x, y) = (4, 0). (Hint: Start by showing y is even.)
We can try the same techniques on y
2
= x
3
+ k for other values of k. The next three
examples illustrate some new features that can occur.
Example 3.6. Consider Mordell’s equation with k = 1: y
2
= x
3
+ 1. (Don’t confuse this
with y
2
= x
3
− 1, which is in Theorem 3.2.) There are several obvious integral solutions:
(x, y) = (−1, 0), (0, ±1), and (2, ±3).
We will use unique factorization in Z to try to show these are the only integral solutions.
4
This will need a lot more work than our use of unique factorization in Z to study y
2
= x
3
+16
in Theorem 3.1.
Rewrite the equation y
2
= x
3
+ 1 in the form
x
3
= y
2
− 1 = (y + 1)(y − 1).
The integers y + 1 and y − 1 differ by 2, so (y + 1, y − 1) is either 1 or 2.
Case 1: y is even. Then y + 1 and y −1 are both odd, so (y + 1, y −1) = 1. (That is, two
consecutive odd integers are always relatively prime.) Since y + 1 and y −1 have a product
that is a cube and they are relatively prime, unique factorization in Z tells us that they
are both cubes or both the negatives of cubes. The negative of a cube is also a cube (since
−1 = (−1)
3
), so y + 1 and y − 1 are both cubes:
y + 1 = a
3
, y − 1 = b
3
.
Subtracting, we have a
3
−b
3
= 2. Considering how cubes spread apart, the only cubes that
differ by 2 are 1 and −1. So a
3
= 1 and b
3
= −1, meaning a = 1 and b = −1. Therefore
y + 1 = 1, so y = 0 and x = −1. The integral solution (−1, 0) of y
2
= x
3
+ 1 is the only
one where y is even.
Case 2: y is odd, so x is even. We expect to show that the only such integral solutions
are (0, ±1) and (2, ±3). Since y +1 and y−1 are both even and differ by 2, (y +1, y −1) = 2.
Either y ≡ 1 mod 4 or y ≡ 3 mod 4. Since (x, y) is a solution if and only if (x, −y) is a
solution, by negating y if necessary we may assume y ≡ 1 mod 4. Then y + 1 ≡ 2 mod 4
and y − 1 ≡ 0 mod 4. Dividing the equation x
3
= y
2
− 1 by 8, we have
x
2
3
=
y + 1
2
·
y − 1
4
.
The two factors on the right are relatively prime, since y+1 and y−1 have greatest common
factor 2 and we have divided each of them by a multiple of 2. Since the product of (y −1)/2
and (y + 1)/4 is a cube and the factors are relatively prime, each of them is a cube:
y + 1
2
= a
3
,
y − 1
4
= b
3
with integers a and b. (Actually, at first we can say (y + 1)/2 and (y − 1)/4 are cubes up
to sign, but −1 = (−1)
3
so, as before, we can absorb a sign into a and b if signs occur.)
Solving each equation for y,
(13) 2a
3
− 1 = y = 4b
3
+ 1,
4
In fact, these are the only rational solutions of y
2
= x
3
+ 1. That is due to Euler [5, Theorem 10]. A
proof by descent is at https://kconrad.math.uconn.edu/blurbs/ugradnumthy/descentbyeuler.pdf.
EXAMPLES OF MORDELL’S EQUATION 9
so a
3
−2b
3
= 1. We can spot right away two integral solutions to a
3
−2b
3
= 1: (a, b) = (1, 0)
and (a, b) = (−1, −1). In the first case, using (13) we get y = 1 (so x = 0) and in the second
case we get y = −3 (so x = 2). We have found two integral solutions to y
2
= x
3
+ 1 when
y ≡ 1 mod 4: (0, 1) and (2, −3). Negating y produces the two solutions (0, −1) and (2, 3)
where y ≡ 3 mod 4.
Therefore if a
3
− 2b
3
= 1 has no integral solution (a, b) besides (1, 0) and (−1, −1), the
equation y
2
= x
3
+1 has no integral solutions besides the five we know.
5
That the only inte-
gral solutions of a
3
−2b
3
= 1 are (1, 0) and (−1, −1) is a special case of the following general
theorem of Delaunay and Nagell [3, pp. 223–226], [7, Sect. VII.3], [9, Sect. 3-9]: for each
nonzero integer d, the equation a
3
−db
3
= 1 has at most one integral solution (a, b) with b 6=
0. The cases d > 0 and d < 0 are equivalent since the exponent is odd: a
3
−db
3
= a
3
+d(−b)
3
.
Proving the Delaunay–Nagell theorem, even for the special case d = 2, introduces many
new complications (a proof of this special case, using p-adic analysis, is in [15, pp. 34–35]
6
or
see https://kconrad.math.uconn.edu/blurbs/gradnumthy/x3-2y3=1.pdf), so we omit
a proof and refer the reader to the indicated references.
Example 3.7. Consider y
2
= x
3
− 5. We have already seen in Theorem 2.2 that this
equation has no integral solutions by a method that only uses calculations in Z. Let’s try
to show there are no integral solutions using factorizations in Z[
√
−5].
Start with a parity check. If x is even then y
2
≡ −5 ≡ 3 mod 8, but 3 mod 8 is not a
square. Therefore x is odd, so y is even.
Write the equation as
x
3
= y
2
+ 5 = (y +
√
−5)(y −
√
−5).
Suppose δ is a common factor of y +
√
−5 and y −
√
−5. First of all, N(δ) divides y
2
+ 5,
which is odd. Second of all, since δ divides (y +
√
−5) − (y −
√
−5) = 2
√
−5, N(δ) divides
N(2
√
−5) = 20. Therefore N(δ) is 1 or 5. If N(δ) = 5 then 5 | (y
2
+ 5), so 5 | y. Then
x
3
= y
2
+5 ≡ 0 mod 5, so x ≡ 0 mod 5. Now x and y are both multiples of 5, so 5 = x
3
−y
2
is a multiple of 25, a contradiction. Hence N(δ) = 1, so δ is a unit. This shows y +
√
−5
and y −
√
−5 have no common factor in Z[
√
−5] except for units.
Since y +
√
−5 and y −
√
−5 are relatively prime and their product is a cube, they are
both cubes (the units in Z[
√
−5] are ±1, which are both cubes). Thus
y +
√
−5 = (m + n
√
−5)
3
for some integers m and n, so
y = m
3
− 15mn
2
= m(m
2
− 15n
2
), 1 = 3m
2
n − 5n
3
= n(3m
2
− 5n
2
).
From the second equation, n = ±1. If n = 1 then 1 = 3m
2
− 5, so 3m
2
= 6, which has no
integral solution. If n = −1 then 1 = −(3m
2
− 5), so 3m
2
= 4, which also has no integral
solution. We appear to have recovered the fact that y
2
= x
3
− 5 has no integral solutions.
Alas, there is an error in Example 3.7. When we wrote certain numbers in Z[
√
−5] as
cubes, we were implicitly appealing to unique factorization in Z[
√
−5], which is in fact false.
5
The converse is true too: every integral solution of a
3
− 2b
3
= 1 leads to the integral solution (x, y) =
(2ab, 4b
3
+ 1) of y
2
= x
3
+ 1, so if y
2
= x
3
+ 1 only has the five integral solutions we know then from
a
3
− 2b
3
= 1 we must have (2ab, 4b
3
+ 1) = (0, 1) or (2, −3) since these are the only (x, y) with y ≡ 1 mod 4,
and from this it easily follows that (a, b) = (1, 0) and (−1, −1).
6
In fact, (1, 0) and (−1, −1) are the only rational solutions of x
3
− 2y
3
= 1, a result first due to Euler [6,
Part II, Sect. II, § 247].
10 KEITH CONRAD
A counterexample to unique factorization in Z[
√
−5] is 6 = 2 · 3 = (1 +
√
−5)(1 −
√
−5).
That doesn’t mean the numbers in Z[
√
−5] that we wanted to be cubes might not be cubes,
but our justification for those conclusions is certainly faulty. It is true in Z[
√
−5] that if
αβ is a cube and α and β are relatively prime then α and β are both cubes, but to explain
why requires new techniques to circumvent the lack of unique factorization.
Example 3.8. Consider y
2
= x
3
− 26. Two obvious integral solutions are (3, ±1). Let’s
use factorizations in Z[
√
−26] to see if (3, ±1) are the only integral solutions.
If x is even then y
2
≡ −26 ≡ 6 mod 8, but 6 mod 8 is not a square. Therefore x is odd,
so y is odd too. Rewrite x
3
= y
2
+ 26 as x
3
= (y +
√
−26)(y −
√
−26). Let δ be a common
factor of y+
√
−26 and y−
√
−26 in Z[
√
−26]. Then N(δ) divides y
2
+26, which is odd. Also
δ divides the difference (y +
√
−26) −(y −
√
−26) = 2
√
−26, so N(δ) divides 4 · 26 = 8 · 13.
Since N(δ) is odd, we see that N(δ) is 1 or 13. There is no element of Z[
√
−26] with norm
13, so N(δ) = 1. Therefore δ = ±1, so y +
√
−26 and y −
√
−26 have only ±1 as common
factors.
If we assume Z[
√
−26] has unique factorization, then since y +
√
−26 and y −
√
−26
multiply to a cube and they have only ±1 as common factors, each of them is a cube. Write
y +
√
−26 = (m + n
√
−26)
3
,
so
y = m
3
− 78mn
2
= m(m
2
− 78n
2
), 1 = 3m
2
n − 26n
3
= n(3m
2
− 26n
2
).
The second equation tells us n = ±1. If n = 1 then 1 = 3m
2
−26, so 3m
2
= 27, which tells
us m = ±3. Then y = (±3)(9 − 78) = ±207 and x
3
= 207
2
+ 26 = 42875 = 35
3
, so x = 35.
We have discovered new integral solutions to y
2
= x
3
− 26, namely (x, y) = (35, ±207). If
n = −1 then 1 = −(3m
2
− 26), so 3m
2
= 25, which has no integral solutions.
Having looked at both possible values of n, we discovered two unexpected integral so-
lutions, but we missed the obvious integral solutions (3, ±1)! How could that happen?
The reason is that our argument was based on the assumption of unique factorization in
Z[
√
−26], but there is not unique factorization in Z[
√
−26]. A counterexample is
27 = 3 · 3 · 3 = (1 +
√
−26)(1 −
√
−26).
It is true that the only integral solutions to y
2
= x
3
− 26 are (3, ±1) and (35, ±207), but a
valid proof has to get around the lack of unique factorization in Z[
√
−26].
4. Rational solutions
For k ∈ Z, if we consider rational solutions to y
2
= x
3
+ k instead of integral solutions,
the situation gets much more complicated. First of all, there could be rational solutions
even if there are no integral solutions. For instance, y
2
= x
3
+ 11 has no integral solutions
by Theorem 2.3, but this equation has the rational solution (x, y) = (−7/4, 19/8). In fact,
y
2
= x
3
+11 has infinitely many rational solutions. Second of all, sometimes the only rational
solutions are the integral solutions, but proving that is much harder than determining all
the integral solutions.
To emphasize the distinction between classifying integral and rational solutions, consider
y
2
= x
3
+16. We proved the only integral solutions are (0, ±4) in Theorem 3.1. This does not
tell us whether there are rational solutions of y
2
= x
3
+16 that are not integral. It turns out
there are no further rational solutions, and here is an application of that. If a
3
+ b
3
= c
3
for
nonzero integers a, b, and c, then the nonzero rational numbers (x, y) = (4bc/a
2
, 4+8(b/a)
3
)
satisfy y
2
= x
3
+ 16. (I found this choice in [2], and moving terms around in the equation
EXAMPLES OF MORDELL’S EQUATION 11
a
3
+ b
3
= c
3
leads to other rational solutions of y
2
= x
3
+ 16. What do you get from
c
3
+ (−b)
3
= a
3
?) Proving the only rational solutions to y
2
= x
3
+ 16 are (0, ±4) would
force x = 0, but x = 4bc/a
2
6= 0, so knowing the only rational solutions of y
2
= x
3
+ 16 are
(0, ±4) implies Fermat’s Last Theorem for exponent 3.
The following table describes all the integral solutions for the cases of Mordell’s equation
we have looked at. (The examples k = 1 and −26 were not fully justified above.)
k Z-solutions of y
2
= x
3
+ k
1 (−1, 0), (0, ±1), (2, ±3)
−1 (1, 0)
−2 (3, ±5)
−4 (2, ±2), (5, ±11)
−5 None
6 None
−6 None
7 None
11 None
16 (0, 4), (0, −4)
−24 None
−26 (3, ±1), (35, ±207)
45 None
46 None
In each case there are finitely many integral solutions, and y
2
= x
3
+ k has finitely many
integral solutions for every nonzero k in Z. If we look at rational solutions, then we might
not get anything new, but we could get a lot that is new. See the next table.
k Q-solutions of y
2
= x
3
+ k
1 (−1, 0), (0, ±1), (2, ±3)
−1 (1, 0)
−2 Infinitely many
−4 Infinitely many
−5 None
6 None
−6 None
7 None
11 Infinitely many
16 (0, 4), (0, −4)
−24 None
−26 Infinitely many
45 None
46 Infinitely many
The equations above that have more rational solutions than integral solutions are y
2
=
x
3
− 2, y
2
= x
3
− 4, y
2
= x
3
+ 11, y
2
= x
3
− 26, and y
2
= x
3
+ 46. Examples of rational
solutions to these equations that are not integral solutions are in the following table.
k −2 −4 11 −26 46
Q-soln
129
100
,
383
1000
106
9
,
1090
27
−
7
4
,
19
8
705
4
,
18719
8
−
7
4
,
51
8
12 KEITH CONRAD
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¨
Ofver diophantiska ekvationen x
n
+ y
n
= z
n
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[13] H. N. Shapiro, “Introduction to the Theory of Numbers,” Wiley, New York, 1983.
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