MHB Would the equation have at least one solution?

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The discussion centers on proving that the diophantine equation y² = x³ + 7 has no solutions. It establishes that if (x₀, y₀) were a solution, x₀ must be odd, leading to contradictions when analyzing the equation modulo 4 and modulo 8. The analysis shows that y² + 1 cannot be divisible by any prime of the form 3 mod 4, which contradicts the conditions derived from the equation. The conclusion is that the equation cannot have any integer solutions, supported by multiple proofs and modular arithmetic reasoning. The overall consensus is that the equation y² = x³ + 7 has no solutions.
evinda
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Hi! (Cool)

In order to show that the diophantine equation $y^2=x^3+7$ has no solution, we do the following:

If the equation would have a solution, let $(x_0,y_0)$, $y_0^2=x_0^3+7$, then $x_0$ is odd.

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

$(x_0-1)^2+3 \in \mathbb{N} \text{ and } (x_0-1)^2+3>1$.

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

So, $(x_0-1)^2+3$ has at least one prime divisor of the form $p \equiv 3 \pmod 4$, so:

$$(x_0-1)^2+3 \equiv 0 \pmod p, \text{ where } p \equiv 3 \pmod 4$$

$$ \Rightarrow y_0^2+1 \equiv 0 \pmod p \Rightarrow y_0^2 \equiv -1 \pmod p$$

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

Therefore:

$$y_0^2 \equiv -1 \pmod p \text{ has no solution}.$$If we would conclude from this relation: $ \left ( \frac{-1}{p} \right )=1$ that $p \equiv 3 \pmod 4$, would we conclude that the diophantine equation has at least one solution? (Thinking)
 
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That is not necessary, no. Even if $-1$ was a quadratic residue modulo $p$, the equation would have been only true modulo $p$. (For example, $2 = 5$ modulo $3$, but $2 \neq 5$ in general :p)
 
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Not entirely on topic, but I'll add another proof I am familiar with.

Assume that $x$ is even. Then $x^3 = 0 \pmod{8}$. Thus $y^2 = x^3 + 7 = 7 = -1 \pmod{8}$, but this is impossible, hence $x$ is odd. Thus $x$ is either $1$ or $3$ modulo $4$.

$$y^2 + 1 = x^3 + 8 = (x + 2)(x^2 - 2x + 4)$$

$y^2 + 1$ is not divisible by any prime $3 \bmod 4$, thus the right hand side is also not divisible by any prime $3 \bmod 4$. In particular, $x + 2$ is not $3 \bmod 4$ (this is because any $3 \bmod 4$ integer must have at least one $3 \bmod 4$ prime factor) and thus is $1 \bmod 4$. Hence $x = 3 \mod 4$ in which case $x^2 - 2x + 4 = 9 - 6 + 4 = 7 = 3 \pmod{4}$ which implies there is a prime factor $p = 3 \pmod{4}$ dividing $x^2 - 2x + 4$, a contradiction $\blacksquare$
 
mathbalarka said:
Not entirely on topic, but I'll add another proof I am familiar with.

Assume that $x$ is even. Then $x^3 = 0 \pmod{8}$. Thus $y^2 = x^3 + 7 = 7 = -1 \pmod{8}$, but this is impossible, hence $x$ is odd. Thus $x$ is either $1$ or $3$ modulo $4$.

$$y^2 + 1 = x^3 + 8 = (x + 2)(x^2 - 2x + 4)$$

$y^2 + 1$ is not divisible by any prime $3 \bmod 4$, thus the right hand side is also not divisible by any prime $3 \bmod 4$. In particular, $x + 2$ is not $3 \bmod 4$ (this is because any $3 \bmod 4$ integer must have at least one $3 \bmod 4$ prime factor) and thus is $1 \bmod 4$. Hence $x = 3 \mod 4$ in which case $x^2 - 2x + 4 = 9 - 6 + 4 = 7 = 3 \pmod{4}$ which implies there is a prime factor $p = 3 \pmod{4}$ dividing $x^2 - 2x + 4$, a contradiction $\blacksquare$

We know that $x$ is odd, so it is of the form $2k+1$.

Therefore:

$$x+2=2k+1+2=2k+3$$

  • $k=2m+1: x+2=2(2m+1)+3=4m+5 \equiv 1 \pmod 4 \Rightarrow x \equiv 3 \pmod 4$
  • $ k=2m: x+2=4m+3 \equiv 3 \pmod 4 \Rightarrow x \equiv 1 \pmod 4$

How do we conclude that $x \equiv 3 \pmod 4$ ? (Thinking)
 
evinda said:
How do we conclude that $x \equiv 3 \pmod 4$ ?(Thinking)

Right, so you have (correctly (Yes)) concluded that $x$ is either $1$ or $3$ modulo $4$.

Now, assume $x = 1 \pmod 4$. Then $x + 2 = 3 \pmod 4$. In that case, $y^2 + 1$ is divisible by some number $3 \pmod 4$. The fact I have used for contradiction is

Claim : Any integer of the form $a^2 + b^2$ is not divisible by any integer $n$ of the form 3 mod 4.

Can you try to prove this claim?
 
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