# $x^3+48=y^4$ does not have integer (?) solutions

How does one find all positive integer solutions to the equation $x^3+48=y^4$?

• I tried factoring the equation as $(y^2 - 4\sqrt{3})(y^2 + 4\sqrt{3}) = x^3$ in $\mathbb{Z}[\sqrt{3}]$ and arrived at a contradiction by assuming that $y^2 - 4\sqrt{3}$ is a cube in $\mathbb{Z}[\sqrt{3}]$. The problem is that I'm not sure about one assumption I made along the way and I can't seem to either justify it or realize that it is nonsense. So I'm at least leaving the comment here just in case someone realizes that it can be proved in this way or at least tell me that I'm wrong. – Adrián Barquero Apr 2 '11 at 5:10
• @Adrián You would have to tell us what exactly you are unsure about. But just a word of caution: it is not enough to show that $y^2-4\sqrt{3}$ is not a cube. It may be a cube times a unit, of which there are infinitely many in your ring. But there are ways of getting round that: the units form an abelian group of rank one and you only need to consider the 0th, 1st, and 2nd power of its generator, which is $2+\sqrt{3}$. So it can be done, but is slightly more cumbersome than meets the eye at first glance. – Alex B. Apr 2 '11 at 5:20
• @Alex Thank you. I just posted my attempt as a community wiki "answer". I totally forgot about the units. I'm just starting to learn Algebraic Number Theory this semester so I'm still too naive with this sort of arguments. – Adrián Barquero Apr 2 '11 at 5:25
• @Adrián Thank you. I hope you don't mind, I have sketched the missing steps in your answer. – Alex B. Apr 2 '11 at 6:01
• @Alex You don't have to thank me at all. I'm the one who has to say thank you. I think I learned a lot from this. – Adrián Barquero Apr 2 '11 at 6:06

I'm posting this as a community wiki "answer" because I'm not sure it works and the explanation wouldn't show up very well in the comments section I guess. If we factor the equation in $\mathbb{Z}[\sqrt{3}]$ we get $(y^2 - 4\sqrt{3})(y^2 + 4\sqrt{3}) = x^3$. Thus if $y^2 - 4 \sqrt{3} = (a + b\sqrt{3})^3$ for some $a + b\sqrt{3} \in \mathbb{Z}[\sqrt{3}]$ then we get

$$y^2 - 4 \sqrt{3} = a^3 + 3a^2b\sqrt{3} + 9ab^2 + 3b^3 \sqrt{3}$$

so that we must have $3a^2 b + 3b^3 = -4$ and since $3$ does not divide $4$ we get a contradiction. The problem is that I need to justify somehow that $y^2 - 4 \sqrt{3} = (a + b\sqrt{3})^3$ and although I know that $\mathbb{Z}[\sqrt{3}]$ is a unique factorization domain, I would need the two factors $y^2 - 4 \sqrt{3}$ and $y^2 + 4 \sqrt{3}$ to be coprime I think, and I can't seem to justify it or disprove it.

Edit (Alex): Suppose that we have an integer solution to $x^3 = y^4 - 48$.

Claim: The two factors $y^2\pm 4\sqrt{3}$ are coprime in $\mathbb{Z}[\sqrt{3}]$. Indeed, let $\alpha = v+\sqrt{3}w$ divide both factors. Then, $\alpha$ also divides their difference, which is $8\sqrt{3}$. Without loss of generality assume that $\alpha$ is irreducible. Then, $\alpha$ divides 2 or is (up to a unit) equal to $\sqrt{3}$. Since $\alpha$ also divides $x^3\in \mathbb{Z}$, we deduce that $x^3$ is divisible by either 2 or 3 (or both). This implies that $y$ must also be divisible by either 2 or 3. But then, $x^3$ is also divisible by either 8 or 27 (or both), since it is a cube, and $y^4$ is divisible by 16 or 81. Now, the latter is impossible, since $y^4 - 48 \equiv -3\neq 0\pmod{9}$. The former is also impossible, since it would imply that $x^3=y^4 - 48$ is divisible by 16, so $x$ is divisible by 4, so $x^3$ is divisible by 64 and you would obtain a contradiction modulo 64 as above.

This implies that $y^2+4\sqrt{3}$ is a cube times a unit in $\mathbb{Z}[\sqrt{3}]$, $y^2+4\sqrt{3}=u\cdot(a+\sqrt{3}b)^3$. The units in this ring form an abelian group with torsion $\pm 1$ and with free part of rank one, generated by $2+\sqrt{3}$. The torsion are all cubes, so we can incorporate them into the bracket. Of course, cubes of a non-torsion unit can also be incorporated into the bracket, so it suffices to show that $y^2+4\sqrt{3}\neq (2+\sqrt{3})^{\delta}\cdot(a+\sqrt{3}b)^3$ for $\delta\in{0,1,2}$. This is done through case by case analysis, Adrián has already done the $\delta=0$ case.

• I finished the analysis of the two remaining cases for $\delta = 1$ and $\delta = 2$ by working modulo $3$. I'll try to edit the answer tomorrow when I have time. – Adrián Barquero Apr 2 '11 at 7:10

One way would be to note that a solution of this equation would also give a solution of the equation $x^3 + 48 = y^2$. That's an elliptic curve of rank 1 and with trivial torsion subgroup. There are bounds on the heights of the integral points on a Weierstrass model of an elliptic curve in terms of the coefficients. You can find a generator for the Mordell-Weil group and using the height bounds find all the integral points. In this case, the only point will be $(x,y) = (1,7)$. Since 7 is not a square, you deduce that the original equation has no integral solutions. Of course, I didn't actually compute the bounds, but asked Magma, which does it automatically. Here is the code:

> P<x>:=PolynomialRing(Integers());
> E:=EllipticCurve(x^3 + 48);
> Rank(E);
1
> TorsionSubgroup(E);
Abelian Group of order 1
> IntegralPoints(E);
[ (1 : 7 : 1) ]
[ <(1 : 7 : 1), 1> ]

• Well, don't forget $(1,-7)$. (Of course $-7$ is not a square either...) Compare with my very similar recent answer, where I pointed out that IntegralPoints does not return Mordell-Weil inverses. (Since the elliptic curve there was not in short Weierstrass form, the additional integral points were not so obvious...) – Pete L. Clark Apr 2 '11 at 15:24
• @Pete Thanks for pointing that out. I have already seen your answer, but decided it wasn't worth bumping this post just for the minus sign. – Alex B. Apr 2 '11 at 15:58
• Thanks for the solutions. :) But is there any simple way to solve this problem? e.g., we may rewrite it as $x^3+64=y^4+16$, and we may use the same method as we use in solving $x^3+23=y^2$. (I mean rewriting as $x^3+27 = y^2 +4$). I really appreciate such a nice solution. – Amir Hossein Apr 2 '11 at 18:14

I realize this is a massive revive, but here is a solution without any powerful theorems about elliptic curves.

We have $x^3 + 64 = y^4 + 2^4$. It is easy to show all (odd) primes dividing the RHS are $1 \pmod{8}$. Now, looking modulo $16$ we deduce either $x \equiv 1 \pmod{16}$ and $y$ is odd or $x,y$ are both even. For the second case if $v_2(x) = 1$ we derive a contradiction because $v_2(LHS) = 3$ while $v_2(RHS) \ge 4$. If $v_2(x) \ge 2$ we have $v_2(LHS) \ge 6$. Thus we must have $v_2(y) = 1$ for $v_2(y^4 + 16) > 4$. Writing $y = 2k$ we need $v_2(k^4 + 1) \ge 2$. But this is absurd since $v_2(k^4 + 1) \le 1$ by checking modulo $4$. Thus it follows $x \equiv 1 \pmod{16}$.

But now utilize $x^3 + 64 = (x+4)(x^2 - 4x + 16)$. $x+4$ is $5 \pmod{8}$, which is absurd so we are done.

• What's $v_2$??? – Graphth Oct 29 '12 at 20:14
• Sorry, I realize this is slightly unconventional notation. It denotes what most people denote as $\text{ord}_p(x)$, the greatest exponent of $p$ which divides $x$. – dinoboy Oct 29 '12 at 20:27
• That's fine, just make sure you explain what it is in your answer. – Graphth Oct 29 '12 at 22:23
• @dinoboy That's not how any people denote $\text{ord}_p(x)$, which is the order of $x$ mod $p$, i.e. the smallest $k\in\Bbb Z^+$ such that $x^k\equiv 1\pmod{p}$. – user236182 Nov 5 '15 at 19:49
• If an odd prime $p$ divides the RHS, then $(-1|p)=1$, i.e. $p\equiv 1\pmod{4}$, but how do you conclude $p\equiv 1\pmod{8}$? – user236182 Nov 5 '15 at 19:50