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How to prove that the only integral solutions to the equation $$y^{2}=x^{3}-1$$ is $x=1, y=0$. I rewrote the equation as $y^{2}+1=x^{3}$ and then we can factorize $y^{2}+1$ as $$y^{2}+1 = (y+i) \cdot (y-i)$$ in $\mathbb{Z}[i]$. Next i claim that the factor's $y+i$ and $y-i$ are co-prime. But i am not able to show this. Any help would be useful. Moreover, i would also like to see different proofs of this question.

Extending Consider the equation $$y^{a}=x^{b}-1$$ where $a,b \in \mathbb{Z}$ and $(a,b)=1$ and $a < b$. Then is there any result which states about the nature of the solution to this equation.

J. M. ain't a mathematician
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Alon Amit's answer is not right, I believe.

$2$ is not prime in the Gaussian integers! $(1+i)$ divides $2$.

The argument upto the fact that the prime $p$ divides $y+i$, $y-i$ and $2i$ is correct.

Since $p|2i$ we consider $p = 1+i$. Now any multiple of $1+i$ is of the form $2x + i2y$ or $2x+1 + i(2y+1)$. Since $y$ is even, $y+i$ cannot be divisible by $1+i$ and thus $y+i$ and $y-i$ are co-prime.

Setting $y+i$ to be a perfect cube (upto units), i.e. $y+i = (xi)^3$, easily gives us $y=0$ and so $x=1, y=0$ is the only solution.

Naweed G. Seldon
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Aryabhata
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If $x$ is even then $x^3$ is divisible by 4, so $x^3-1 \equiv 3 \pmod 4$ and this cannot be a square. Thus $x$ is odd and $y$ is even.

Now, if a prime $p$ divides both $y+i$ and $y-i$ then it divides their difference $2i$. Thus $p=2$ (up to units), but then $p$ divides $y$. That's impossible since it divides $y+i$ as well.

EDIT: I was too hasty in writing this, as pointed out by Moron. The prime 2 ramifies in $\mathbb{Z}[i]$, and is (up to units) the square of $1+i$. Sorry for the confusion.

Alon Amit
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Let $\alpha\in\mathbb{Z}[i]$ be a divisor of $y+i$ and $y-i$. Then $\alpha|2=i(y-i-(y+i))$ and $\alpha|(y-i)(y+i)=x^3$. Since $x$ is odd then $x^3$ is odd and therefore, by Bezout there exist $A,B\in\mathbb{Z}$ such that $Ax^3+2B=1$ and therefore $\alpha|1$ implying $\alpha\in\mathbb{Z}[i]^{\times}$. We conclude that $y+i$ and $y-i$ are coprime.

Potitov06
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Is it really necessary to use complex numbers here? Does this sound reasonable to you or is there a flaw? According to the 7-period for n^2 and n^3-1 (mod 7) LHS=RHS requires both to be 0 mod(7), so that for some n and k y^2=n*7^2 must equal k* 7^3 -1 and therefore 7^2(n-7k)=-1 , impossible, since 7^2>1. I note the table format doesn't come out right - you have to do it yourselves.

  • Welcome to MSE! I realize you don't yet have enough reputation, but this is better left as a comment as it is not explicitly answering the question and posing another one. Regards – Amzoti Apr 09 '13 at 12:52
  • @Mikael Jensen, your reasoning is wrong: it is true that for y^2 and x^3-1 to be equal (mod 7) both must equal 0 (mod 7). This indeed implies that y^2=0 (mod 7) and hence y=0 (mod 7), y=7n and y^2=7^2n^2, but x^3-1=0 (mod 7) does _not_ imply that x^3-1=7^3k^3-1! It implies that x^3=1 (mod 7) and thus x=1,2 or 4 (mod 7). If, for example, x=7k+1, then x^3-1=(7k+1)^3-1 = (7k)^3+3(7k)^2+3(7k), and your argument fails. – Jose Brox Aug 13 '14 at 10:17
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I think I have found a proof for the statement made in the top answer:

Setting $y+i$ to be a perfect cube (upto units), i.e. $y+i = (xi)^3$, easily gives us $y=0$ and so $x=1, y=0$ is the only solution.

Let $y + i = (a + bi)^3 = (a^3 - 3ab^2) + i(3a^2b - b^3),$ then by comparison of coefficients, $1 = 3a^2b - b^3 = b(3a^2 - b^2)$, and therefore $b = \pm 1$.

  • Case 1: $b = 1\colon$ $3a^2 = 2 \Rightarrow a = \sqrt{2/3} \not\in \mathbb Z.$
  • Case 2: $b = -1\colon$ $3a^2 = 0 \Rightarrow a = 0 \Rightarrow y+i = (-i)^3 = i \Rightarrow y = 0 \Rightarrow x = 1.$
Paul W.
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