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I'm trying to find all integer points on an elliptic curve $y^2=x(x^2-2x-3)$ or $y^2=x(x+1)(x-3) $. I guess there are only 3 integer points: $(-1,0), (0,0), (3,0) $. It looks like quite an elementary number theory problem, but I couldn't think up any proof... How could I prove it, or disprove it? Please give me at least some hints.

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Here's an elementary argument. Suppose none of $x$, $x+1$, or $x-3$ is $0$, and that their product is a perfect square. Note that for any prime $p\ge5$, $p$ can divide at most one of $x$, $x+1$, and $x-3$. Thus the only primes that could have an odd power in the prime factorization of any of these three numbers are $2$ and $3$, so each of them has the form $\pm a^2$, $\pm 2a^2$, $\pm 3a^2$, or $\pm 6a^2$ (where $a$ is an integer). Let us say a number has a "spare factor" of a prime if that prime appears with an odd exponent in its prime factorization. In order for $x(x+1)(x-3)$ to be a perfect square, only $x$ or $x-3$ can have a spare factor of $3$ (in which case both of them do), and only $x+1$ or $x-3$ can have a spare factor of $2$ (in which case both of them do).

Let's now consider cases based on whether there are spare factors of $2$ or $3$. If there are no spare factors of $2$ or $3$, then all three numbers are of the form $\pm a^2$. This is impossible: $x$ and $x+1$ can't both be of this form unless one of them is $0$.

If there are spare factors of $2$ but not spare factors of $3$, then $x+1$ and $x-3$ both have the form $\pm2a^2$. Dividing by $2$, we get two numbers of the form $\pm a^2$ with a difference of $2$, which is only possible if they are $-1$ and $1$. This means $x+1=2$ and so $x=1$. But if $x=1$ then $x(x+1)(x-3)=-4$, which is not a perfect square.

If there are spare factors of $3$ but not spare factors of $2$, then $x$ and $x-3$ both have the form $\pm 3a^2$. Dividing by $3$, we get numbers of the form $\pm a^2$, which is impossible since it would imply $x$ or $x-3$ is $0$.

Finally, suppose there are both spare factors of $3$ and spare factors of $2$. Note that $x$ must be $1$ mod $4$ (if it were $3$ mod $4$, then one of $x+1$ and $x-3$ would be divisible by $4$ and not $8$, and thus would not have a spare factor of $2$). Also, $x$ must have the form $\pm 3a^2$ since it has a spare factor of $3$. Now no number of the form $3a^2$ can be $1$ mod $4$, so $x$ must actually have the form $-3a^2$. But if $a\neq 0$, this would imply all three of $x$, $x+1$, and $x-3$ are negative, so their product is negative and not a square.

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