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Prove there are no $x,y\in\mathbb{Z}$ such that $3x^2=9+y^3$.

Initial proof

Let us assume there are $x,y\in\mathbb{Z}$ that satisfy the equation, which can be rewritten as $$3(x^2-3)=y^3.$$ So, $$3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3.$$ As $3 \mid -3$, it follows that $$3 \mid x^2 \Rightarrow 3^2 \mid x^2.$$ Say $x^2=3^2 \cdot a^2$ for some $a \in \mathbb{Z}$. Then $$\begin{align*}x^2-3 &= 3^2 \cdot a^2 - 3 \\&= 3 \cdot (3 \cdot a^2 - 1)\end{align*}$$ As $3^2 \mid x^2-3$, it follows that $3 \mid 3 \cdot a^2 - 1$. It is obvious that $3 \mid 3 \cdot a^2$, so it follows that $3 \mid -1$, which is false. Therefore, the assumption that there are $x,y\in\mathbb{Z}$ was wrong.

Alternative proof

$$ \begin{align*} 3x^2&=9+y^3\\ 3(x^2-3)&=y^3 \end{align*} $$

So, $3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3 \Rightarrow 3^2 \mid x^2 \Rightarrow 3^2 \not\mid x^2 - 3$ and we are done.

Both proofs are essentially the same, except for the fact that the second proof is much shorter. I think the first proof is much more clear, but it takes a bit longer to write down.

  • Would the second proof suffice in a math competition, or is it too short?
  • This problem was part of the Dutch finals of 1978. I figured it out in under a minute, which is quite unusual for me, so now I wonder: is my proof correct?
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I'd probably start the sequence of implications in the alternative proof with $3\ |\ y \Rightarrow 3^3\ |\ y^3 \Rightarrow \ldots$ but other than that, the proof looks okay to me -- and it would suffice in the math competitions I know of. Of course, every competition might have its own set of rules of what would be acceptable. – Peter Košinár Jun 13 '13 at 19:17
@PeterKošinár You're right, I would definitely write that down in a contest, and I've added it to the proof. – Tim Vermeulen Jun 13 '13 at 19:29
In the "alternative proof", the implication $3^2\mid x^2-3\implies 3^2\mid x^2$ is dubious since $3^2$ is not a divisor of $3$. The "initial proof" is correct. – Did Jun 14 '13 at 5:22
@Did The implication is based on the fact that $3^2 \mid x^2 - 3 \Rightarrow 3 \mid x^2 - 3 \Rightarrow 3 \mid x^2 \Rightarrow 3^2 \mid x^2$, but you're right, I probably should not omit it. I was wondering if it would be enough, but it is indeed a bit dubious. – Tim Vermeulen Jun 17 '13 at 19:54
up vote 1 down vote accepted

Your second proof starts without an assumption and this is not a matter of length. The word "Assume" before the first formula would suffice. This oversight would probably be forgiven if your work ended with a contradiction and the consequences which it does not.

Between $3^2\mid x^2-3$ and $3^2 \mid x^2$, you should add at least $3\mid x$ and ideally $3\mid x^2-3 \Rightarrow 3\mid x^2 \Rightarrow 3\mid x$.

You are highly likely to lose some points on that step, although it is not certain that you would.

The difference between you first and second version is NOT the shortness and lack of words, but the fact that you left out that step.

The second issue is that you do not say in your second proof that there are no integer solutions. You are certain to lose points on that. So, for example, you could circle the two contradictory assertions, make arrows to the next line and write "Contradiction. There are no solutions."

It is not an issue of length, but whether every step is addressed in some way.

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Thanks. I would never write a proof like my second one here on an actual contest or exam, but I've seen proofs to other problems that don't go into much detail at all (which I tried to mimic with my second proof). – Tim Vermeulen Aug 19 '13 at 15:51

Suppose there is a solution to $3x^2 = 9 + y^3$.

Then $3$ divides the RHS hence $3|y^3$, so $3|y$. Let $y = 3a$ for some integer $a$.

Then $3x^2 = 9 + 27a^3$. Hence $9 | 3x^2$ and so $3|x^2$, implying $3|x$. Let $x = 3b$ for some integer $b$.

Then $27b^2 = 9 + 27a^3$, and after cancelling $9$'s we get $3b^2 = 1 + 3a^2$ or $3(b^2 - a^2) = 1$, which clearly has no solutions since the LHS is divisible by $3$.

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Assume that the expression has integer solutions, or atleast one for that matter:

Work in modulus $3$, if:


$y^3$ must be a multiple of $3$. Then we can set $y=3k$ where $k$ is an integer. Now we have:


Dividing both sides by $3$ gives us $9k^3=x^2-3$ Taking $-3$ to the other side shows us that $x^2$ is a mutiple of $3$. Thus $x$ is a multiple of $3$. Then we can set $x=3t$ where $t$ is an integer. This gives us:

$$9k^3+3=9t^2$$ Dividing both sides by $3$ again leaves us with this: $$3k^3+1=3t^2$$ Which is impossible because $3t^2$ is a multiple of $3$ and $3k^3+1$ is not.

This is a contradiction. Thus the proof is complete.

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Could you please use MathJax to typeset your work? It is pretty easy to learn - just go here:… – Antonios-Alexandros Robotis Feb 24 at 7:16

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