How to find ALL pairs of positive integers $(x,y)$ such that the difference in their squares is a perfect cube and the difference in their cubes is a perfect square.
i.e., Positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$?

Finding infinite number of pairs is no problem, as in:

$( 2^{6j+1} \cdot 3^{6k} \cdot 5 , 2^{6j+1} \cdot 3^{6k+1} )$ for any integral $j,k \geq 0$

But how would you determine the exhaustive list?

  • $\begingroup$ Your example seems to arise from positive integers $r,s$ such that $(r^3 - s^3)/(r^2 - s^2)$ is a perfect square, as appropriate common multiples of $r,s$ then give $x,y$ to satisfy your equations. E.g. $(5^3 - 3^3)/(5^2 - 3^2) = 49$. $\endgroup$ – hardmath Jul 23 '11 at 13:46
  • $\begingroup$ @hardmath: That is true for my example, but my example does not account for all possibilities and in others that does not hold; e.g., for pairs of the form: $( 2 \cdot 11^{(6k+2)} \cdot 37 , 11^{(6k+2)} \cdot 47 )$ $\endgroup$ – ScottT Jul 23 '11 at 15:05

$ x^2 - y^2 = a^3 $ is fairly trivial, because clearly for every integer solution there are integers $ p, q, r, s $ with $p$ and $q$ squarefree and coprime such that $ x + y = p q^2 r^3 $ and $ x - y = p^2 q s^3 $, so that:

$ 2 x = p q (q r^3 + p s^3) $

$ 2 y = p q (q r^3 - p s^3) $

$ a = p q r s $

and these satisfy the equation identically, with $x, y$ integers iff either $p q$ is even or $p, q, r, s$ all odd.

(or $r, s$ both even, although for "reduced" solutions with no integer $t > 1$ such that t^3 divides $x, y$ and t^2 divides $a$ we can assume that $gcd(r, s) = 1$, which we do hereafter ..)

Plugging these relations into $(2 x)^3 - (2 y)^3 = 8 b^2 $ gives $ p^4 q^3 (3 q^2 r^6 s^3 + p^2 s^9) = 4 b^2 $.

Noting that $p$ and $q$ are coprime and squarefree we then conclude $2 b = p^2 q^2 s B$ for some integer $B$, so that $ s (3 q^2 r^6 + p^2 s^6) = q B^2 $.

From this we see that $q$ divides $p^2 s^7$ and hence, since $gcd(p, q) = 1$ and $q$ is squarefree, that $q$ divides $s$.

Thus with $s = q S$ and $B = q C$ we obtain $ S (3 r^6 + p^2 q^4 S^6) = C^2 $.

Then, taking $S = u v^2$ with $u$ squarefree, we conclude that $C = u v D$ for some integer $D$ so that finally $3 r^6 + (p q^2 u^3 v^6)^2 = u D^2 $

In this $u$ divides $3 r^6$; but since $gcd(r, s) = 1$ and $u$ divides $s$ we conclude that $u$ divides 3.

Thus, since $u$ > 0, we must have $u = 1$ or $u = 3$, which reduces the problem to one of the following respectively:

$3 r^6 + M^2 = N^2$ ($u = 1$)

$ r^6 + 3 M^2 = N^2$ ($u = 3$)

Now $X^2 + 3 Y^2 = Z^2$ has general integer solution $X, Y, Z = k(m^2 - 3 n^2), 2 k m n, k(m^2 + 3 n^2)$ with $gcd(m, n) = 1$ and $m + n$ odd.

So the cases require either $2 k m n = r^3$ or $k(m^2 - 3 n^2) = r^3$, each of which is trivial by suitable choice of $k$ (although arguably if you want explicit values of m, n the second is not so trivial).

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All the possible positive integers $x,y$ such that $x^3 - y^3$ is a perfect square and $x^2 - y^2$ is a perfect cube can be described in terms of reducing $x/y$ to lowest terms $r/s$.

That is, suppose $x = cr$ and $y = cs$ with $r,s$ coprime. Since $x \gt y$, we choose $r > s$.

First we can uniquely factor $r^3 - s^3 = z^2 t$ where $t$ is squarefree.

Furthermore we can uniquely factor $r^2 - s^2 = u^3 v^2 w$ where $v,w$ are squarefree and coprime.

We show that $c$ is uniquely determined up to an arbitrary sixth power factor. That is:

$$c = (c_1 c_2^2 c_3^3 c_4^4 c_5^5)*d^6$$

where the factors $c_i$ are squarefree, pairwise coprime, and to be determined below, and $d$ is any positive integer.

Substituting for $x,y$ in $x^3 - y^3$ and $x^2 - y^2$, we get these expressions:

$$ (cr)^3 - (cs)^3 = (c_1^3 c_2^6 c_3^9 c_4^{12} c_5^{15}) d^{18} (z^2 t) $$ $$ (cr)^2 - (cs)^2 = (c_1^2 c_2^4 c_3^6 c_4^{8} c_5^{10}) d^{12} (u^3 v^2 w) $$

Removing evident squares from the first right-hand side, resp. cubes from the second, we get these conditions that must be satisfied:

$(i) \;\; c_1 c_3 c_5 t$ is a perfect square

$(ii) \;\; c_1^2 c_2 c_4^2 c_5 v^2 w$ is a perfect cube

Using coprimality and squarefreeness of the various factors, the following expressions for the factors $c_i$ are implied:

$c_1 = \gcd(t,w)$

$c_5 = \gcd(t,v)$

$c_3 = t/(c_1 c_5)$

$c_2 = v/c_5$

$c_4 = w/c_1$


We claim that $t = c_1 c_3 c_5$ follows from $c_1 c_3 c_5 t$ being a perfect square, in light of the squarefreeness of each factor and the pairwise coprimality of the $c_i$. For any prime factor of $t$ appears to a first power there (owing to the squarefreeness of $t$) and must appear an odd number of times in $c_1 c_3 c_5$ in order for the product of that with $t$ to be a perfect square. But any prime factor of $c_1 c_3 c_5$ can only occur once because the $c_i$ are pairwise coprime and squarefree. Thus the prime factors of $t$ are distinct and in correspondence with the (distinct) prime factors of $c_1 c_3 c_5$. Their equality follows.

By similar arguments we deduce from $c_1^2 c_2 c_4^2 c_5 v^2 w$ being a perfect cube, in light of the squarefreeness of $v,w$ as well as the $c_i$ and coprimality of $v,w$ (as well as the pairwise coprimality of the $c_i$), that $v = c_2 c_5$ and $w = c_1 c_4$.

Now clearly $c_1 = \gcd(t,w)$ and $c_5 = \gcd(t,v)$. The rest of the expressions, for $c_2,c_3,c_4$, are then forced. QED


For any coprime pair $r \gt s$ we can find a smallest solution of the form $x = cr$ and $y = cs$, and all other solutions where $x/y = r/s$ come from scaling up by an arbitrary sixth power.

The pairs ScottT cites in the Question are among those for which $(r,s) = (10,3)$, and those in a Comment underneath for which $(r,s) = (74,47)$.

To illustrate let's first pick a simple coprime pair, say $(r,s) = (2,1)$.

Then $r^3 - s^3 = z^2 t$ with $t$ squarefree means $t = 7$ and $z = 1$.

Similarly $r^2 - s^2 = u^3 v^2 w$ with $v,w$ squarefree and coprime means $u = 1$, $v = 1$, $w = 3$.

The smallest solution is given by $c = c_1 c_2^2 c_3^3 c_4^4 c_5^5$ where:

$c_1 = \gcd(t,w) = 1$

$c_5 = \gcd(t,v) = 1$

$c_3 = t/(c_1 c_5) = 7$

$c_2 = v/c_5 = 1$

$c_4 = w/c_1 = 3$

Thus the smallest $c = 7^3 * 3^4 = 27783$, which corresponds to $x = 55566$ and $y = 27783$. Verifying:

$$x^3 - y^3 = (r^3 - s^3)c^3 = 7 * (7^3 * 3^4)^3 = 7^{10} * 3^{12}$$

which is a perfect square, and:

$$x^2 - y^2 = (r^2 - s^2)c^2 = 3 * (7^3 * 3^4)^2 = 7^6 * 3^9$$

which is a perfect cube. All other solutions with $x/y = 2/1$ are obtained from this one by scaling with a factor $d^6$.

To reprise ScottT's first example $(r,s) = (5,3)$ gives $t = 2$, $v = 1$, and $w = 2$. Then the smallest $c = 2$ gives $x = 10$ and $y = 6$, for which $x^3 - y^3 = 28^2$ and $x^2 - y^2 = 4^3$.

ScottT's second example $(r,s) = (74,47)$ gives:

$74^3 - 47^3 = 301401 = 549^2$

so $t = 1$, and:

$74^2 - 47^2 = 3267 = 3^3 * 11^2$

so $v = 11$ and $w = 1$. The smallest $c = 11^2$ gives $x = 8954$ and $y = 5687$. Note $x^3 - y^3 = 730719^2$ and $x^2 - y^2 = 363^3$.

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  • $\begingroup$ I'm sorry, what does this actually prove? Does it find all $x,y$ with the desired property? Does it find any $x,y$ with the desired property? $\endgroup$ – Gerry Myerson Jul 24 '11 at 4:17
  • $\begingroup$ @Gerry Myerson: It does find all $x,y$ with the desired property. "Any" $x,y$ with the desired property has the form $(cr,cs)$ where $r \gt s$ are coprime (you choose) and $c$ is an arbitrary sixth power $d^6$ times a number $C(r,s)$ determined by $r$ and $s$ as above. Perhaps an example will make it clear? $\endgroup$ – hardmath Jul 24 '11 at 10:57
  • $\begingroup$ Sorry, still not with you, as I don't see any $C(r,s)$ in your write-up. Are you saying that for any coprime $r,s$ there exists this $C$ such that $x=Cr,y=Cs$ is a solution? $\endgroup$ – Gerry Myerson Jul 24 '11 at 12:41
  • $\begingroup$ OK, so, $C$ is $c$. Very good. Can it ever happen that $c=1$, and $x,y$ are coprime? $\endgroup$ – Gerry Myerson Jul 24 '11 at 12:51
  • $\begingroup$ @Gerry Myerson: Let's say $C(r,s)$ is the smallest $c$ that gives a solution $(cr,cs)$ for coprime $r \gt s$. I'm afraid the construction above sheds almost no light on whether $c = 1$ is possible. By analogy with the van der Poorten/Beal conjecture, your coprime wrinkle looks to be difficult unless we stumble on an example. $\endgroup$ – hardmath Jul 24 '11 at 16:41

Not an answer, just thinking out loud about the question of whether we can ever take $x,y$ relatively prime.

We solve $x^2-y^2=a^3$ by writing $a^3=a_1a_2$ with $a_1,a_2$ of the same parity, then $x-y=a_1$, $x+y=a_2$, $2x=a_2+a_1$, $2y=a_2-a_1$, and we see any common divisor of $a_1,a_2$ other than perhaps $2$ is a common divisor of $x,y$. So we have $a_1=c^3$, $a_2=d^3$ with $c,d$ relatively prime, or else $a_1=2c^3$, $a_2=4d^3$, or else $a_1=4c^3$, $a_2=2d^3$. I'll just look at the first case.

We have $2x=d^3+c^3$, $2y=d^3-c^3$, $c,d$ relatively prime. Then $$8b^2=(2x)^3-(2y)^3=(d^3+c^3)^3-(d^3-c^3)^3=6d^6c^3+2c^9$$ so $(2b)^2=c^3(3d^6+c^6)$. Since $c,d$ are coprime, the only possibilities for $\gcd(c^3,3d^6+c^6)$ are $1$ and $3$. I'll just look at the first case.

Now $c^3$ and $3d^6+c^6$ must be squares, so $c=e^2$ and $3d^6+e^{12}=f^2$. So $$(f-e^6)(f+e^6)=3d^6$$ If $f-e^6,f+e^6$ are coprime (and they must be pretty nearly coprime, if you trace back through to the coprimality of $c,d$), then $f-e^6=g^6$, $f+e^6=3h^6$, or else $f-e^3=3g^6$, $f+e^6=h^6$. Taking, as usual, just the first case, we get $$g^6+2e^6=3h^6$$

That's as far as I go. There is the trivial solution, $e=g=h=1$. That rules out using congruences to show there aren't any solutions, but I suspect there aren't any other coprime solutions. Maybe someone can actually prove this, and clean up the cases I haven't discussed.

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  • $\begingroup$ Maybe the relation $3d^6 + e^{12} = f^2$ would give rational approximations to $\sqrt{3}$ that are "better than one should be." $\endgroup$ – hardmath Jul 26 '11 at 2:00
  • $\begingroup$ @hardmath, I've seen that kind of argument apply to 2-variable equations, but I don't see how to make it go here, with 3 variables. $f^2-3d^6$ isn't small, it's $e^{12}$, which could be of size comparable to $d^6$ and/or $f^2$. $\endgroup$ – Gerry Myerson Jul 26 '11 at 6:23

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