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Can an even square exceed a cube by one?

I can't find any examples so I assume it is false. How would I prove it?

I forgot to add that they must be positive.

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    $\begingroup$ Well, there is $(-1)^3+1=0^2$. (There are no others.) $\endgroup$ – André Nicolas Dec 9 '13 at 3:28
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    $\begingroup$ @JackAidley 1 is not even $\endgroup$ – Ivo Beckers Dec 9 '13 at 8:39
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Here is a proof using elementary methods:

We want to find positive integer solutions to the equation $(2m)^2 = n^3 + 1$. This is equivalent to $$ n^3 = (2m+1)(2m-1).$$ If $p$ is any prime in the prime factorization of $2m+1$ then from the equation above $p$ divides $n^3$. Because $n^3$ is a perfect cube, the exponent of $p$ in the prime factorization of $n^3$ must be divisible by 3. Then since $2m+1$ and $2m-1$ are relatively prime, $2m+1$ must contain all of $n^3$'s factors of $p$. So the exponent of $p$ in the prime factorization of $2m+1$ must be divisible by 3. This is true for every prime factor $p$ of $2m+1$, so $2m+1$ is a perfect cube.

Using a similar argument we see that $2m-1$ must also be a perfect cube.

So we have two perfect cubes whose difference is $(2m+1)-(2m-1) = 2$. But this is impossible, since the difference between two (distinct) positive perfect cubes is at least $2^3 - 1^3 = 7$.

Hence the equation $(2m)^2 = n^3 + 1$ has no positive integer solutions.

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You are looking for positive integer solutions to $(2a)^2 = b^3+1$, or equivalently for $(2a-1)(2a+1)=b^3$.

Now we have two consecutive odd numbers on the LHS, so they are relatively prime (any common factor has to divide $2$, and they are odd). Hence they both must be cubes.

Now it is easy to show that two consecutive cubes $x^3$ and $(x+1)^3$ are separated by $3x^2+3x+1$ which would be more than $2$ for positive $x$.

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If I remember correctly, the elliptic curve $y^2=x^3+1$ only has $(0,\pm 1)$, $(-1, 0)$ and $(2, \pm 3)$ as rational points (not including infinity). The only solution with $y$ even was already mentioned in the comments.

Edit': Thanks to Álvaro Lozano-Robledo for finding the reference, and Keith Conrad for informing me that the result is only quoted in the reference, not actually proven. Nonetheless, I am now sure that the result is true, and have an idea of the proof. Álvaro Lozano-Robledo suggested a strategy using $2$-descent to show that the rank of this curve is $0$, and then using the Nagell-Lutz theorem to find the points of finite order. I will return to this post in some finite amount of time after doing some more reading and see if I can supply the proof.

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  • $\begingroup$ Why?‍‍‍‍‍‍‍‍‍‍‍‍ $\endgroup$ – chubakueno Dec 9 '13 at 4:52
  • $\begingroup$ Why are these the only rational points? That's a great question and I don't have a good answer. I just remember reading (somewhere) that this curve had rank $0$, and that these were the only rational points. Hopefully an expert sees this and can remark on it or give a reference. $\endgroup$ – Dylan Yott Dec 9 '13 at 4:54
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    $\begingroup$ My colleague Keith Conrad has some notes on this here: math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf, see Example 3.6. $\endgroup$ – Álvaro Lozano-Robledo Dec 9 '13 at 14:14
  • $\begingroup$ needs another $y^2 = x^3 \large\bf{+1}$ upswag $\endgroup$ – Stahl Dec 9 '13 at 17:18
  • $\begingroup$ The treatment of integral solutions to $y^2 = x^3 + 1$ in Example 3.6 of the reference Alvaro links to is incomplete. It just turns that problem into the problem of solving $x^3 - 2y^3 = 1$ in integers. $\endgroup$ – KCd Dec 10 '13 at 4:38
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Catalan's conjecture states that there are no two powers (with exponents $\geq2$) of natural numbers at distance $1$ from each other, apart from $8$ and $9$. The conjecture has been proven by Preda Mihailescu in 2002.

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