I am looking at the following exercise:

The diophantine equation $y^2=x^3+7$ has no solution.

If the equation would have a solution, let $(x_0,y_0)$,then $x_0$ is odd.

( If $x_0$ is even, $x_0=2k \Rightarrow x_0^3 \equiv 0 \pmod 8$, therefore $y^2 \equiv 7 \pmod 8$, that is a contradiction, since $y^2 \equiv 0,1,4 \pmod 8$ )

$$y_0^2=x_0^2+7 \Rightarrow y_0^2+1=x_0^3+8=x_0^3+2^3=(x_0+2)(x_0^2-2x_0+4)=(x_0+2)[(x_0-1)^2+3]$$

$$(x_0-1)^2+3 \in \mathbb{N}$$ $$(x_0-1)^2+3>1 , \text{ so it has a prime divisor, let }p$$

It stands $(x_0-1)^2+3 \equiv 3 \pmod 4$

If $p_1 \equiv 1 \pmod 4$ and $p_2 \equiv 1 \pmod 4$, $p_1 \cdot p_2 \equiv 1 \pmod 4$.

Therefore, there is at least one prime divisor of $(x_0-1)^2+3$, of the form $p \equiv 3 \pmod 4$, so $(x_0-1)^2+3 \equiv 0 \pmod p \Rightarrow y_0^2+1 \equiv 0 \pmod p \Rightarrow y_0^2 \equiv -1 \pmod p$

$$Y^2 \equiv -1 \pmod p \text{ has a solution } \Leftrightarrow \left ( \frac{-1}{p}\right )=1 \Rightarrow (-1)^{\frac{p-1}{2}}=1 \Rightarrow \frac{p-1}{2}=2\\ \Rightarrow p \equiv 1 \pmod 4$$

So, $y_0^2 \equiv -1 \pmod p$ has no solution.

Is there also an other way, to prove that the diophantine equation $y^2=x^3+7$ has no solution?


It so happens that the ring $\mathbb{Z}[\sqrt{7}]$ is a UFD, so we can rewrite this as:

$$(y-\sqrt{7})(y+\sqrt{7}) = x^3$$

If $7 \mid y$, we get a contradiction $\pmod{49}$. Likewise, if $2\mid y$, we get a contradiction $\pmod{8}$ (your own reasoning shows this).

Otherwise, $\gcd(y-\sqrt{7}, y+\sqrt{7}) = 1$, implying that both $y-\sqrt{7}$ and $y+\sqrt{7}$ are cubes in $\mathbb{Z}[\sqrt{7}]$.

So there exist $a,b\in\mathbb{Z}$ with $y+\sqrt{7} = (a+b\sqrt{7})^3 = (a^3 + 21ab^2) + (3a^2 b + 7b^3)\sqrt{7}$, so $y=a^3+21ab$, $1=3a^2b + 7b^3$.

The second equation is $b(3a^2 + 7b^2) = 1$. But $3a^2+7b^2$ is never a divisor of $1$, contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.