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EXAMPLES OF MORDELL’S EQUATIONKEITH CONRAD
1. Introduction
The equation y 2 = x3 + k, for k ∈ Z, is called Mordell’s equation1 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 = x3 + 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 = x3 +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 = x3 + 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 = x3 + 7 as y 2 + 1 = x3 + 8, so
(1) y 2 + 1 = x3 + 8 = (x + 2)(x2 − 2x + 4).
Note x2 − 2x + 4 = (x − 1)2 + 3 is at least 3. Since x is odd, (x − 1)2 + 3 ≡ 3 mod 4. Thus
x2 − 2x + 4 has a prime factor p ≡ 3 mod 4: if not, all of its prime factors are 1 mod 4,
so x2 − 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
x2 − 2x + 4 ≡ 3 mod 4. From p | (x2 − 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].
1Also called Bachet’s equation.
2Large 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 x2 − 2x + 4. Since
(as seen above) x is odd and y is even, x3 ≡ x mod 4 (true for all odd x), so reducing
y 2 = x3 + 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 x3 ≤ −8, so x3 + 7 ≤ −1, which contradicts x3 + 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 = x3 − 5 has no integral solutions.
Proof. Assuming there is a solution, reduce modulo 4:
y 2 ≡ x3 − 1 mod 4.
Here is a table of values of y 2 and x3 − 1 modulo 4:
y y2 mod 4 x x3 − 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 x3 − 1 mod 4 is 0, so y is even and x ≡ 1 mod 4.
Then rewrite y 2 = x3 − 5 as
(2) y 2 + 4 = x3 − 1 = (x − 1)(x2 + x + 1).
Since x ≡ 1 mod 4, x2 + x + 1 ≡ 3 mod 4, so x2 + x + 1 is odd. Moreover, x2 + x + 1 =
(x + 1/2)2 + 3/4 > 0, so x2 + x + 1 ≥ 3. Therefore x2 + x + 1 must have a prime factor
p ≡ 3 mod 4 (same reasoning as in the previous proof). Since p is a factor of x2 + 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 = x3 + 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 = x3 + 11 as
(3) y 2 + 16 = x3 + 27 = (x + 3)(x2 − 3x + 9).
The factor x2 −3x+9 is positive (why?), and from x ≡ 1 mod 4 we get x2 −3x+9 ≡ 3 mod 4,
so by the same reasoning as in the proof of Theorem 2.1, x2 − 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 = x3 − 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 x3 = y 2 + 6 ≡ 7 mod 8. Also
x3 ≡ x mod 8 (true for all odd x), so x ≡ 7 mod 8.
Rewrite y 2 = x3 − 6 as
(4) y 2 − 2 = x3 − 8 = (x − 2)(x2 + 2x + 4),
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