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The Mordell equation is the Diophantine equation $$Y^2 = X^3-k \tag{1}$$ where $k$ is a given integer. There is no known single method — elementary or otherwise — to solve equation $(1)$ for all $k$, though there are various ad hoc methods which work on large classes of $k$, and there are [published] elementary solutions for many (though not all) $k$.

For the scope of this discussion, I define “elementary” as pre-college level number theory — definitely excluding imaginary numbers, class numbers, imaginary quadratic fields, "advanced" UFDs, and so on; almost certainly excluding quadratic reciprocity and similar; but definitely including FTA (unique factorization in the regular integers), mathematical induction, polynomial division, simple modular arithmetic, and so on.

Emboldened by my recent completely elementary solution to the Mordell equation for $k=-1$, I am wondering if an elementary proof can be found for every Mordell equation. Since it is impossible to prove a negative, I guess my real question is this:

For any integer or class of integers $k$, are there any obvious impediments to the development of an elementary solution?

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Kieren MacMillan
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    $Y^2 = X^3 + 7$ has no integer solutions. Would you consider this "a solution"? – RghtHndSd Nov 15 '14 at 18:00
  • Yes, by "solution" I mean "proof by which all integer solutions are obtained, or it is shown that there are no integer solutions". So if you can, using only elementary means [as described above], prove that $Y^2 = X^3 + 7$ has no integer solutions, that would be a non-counterexample to my implied conjecture [that all Mordell equations admit a completely elementary "solution"]. – Kieren MacMillan Nov 15 '14 at 18:25
  • @KierenMacMillan One obstruction is that we don't know anything about $k$ except that it's an integer. The elementary soltions found so far rely on the properties of the integer (i.e, if it is a perfect square, perfect cube, it's form is $5x^2+1$ etc.). Btw, do you know where can we get these "elementary" proofs of Mordell Equations published? If so, can you please notify me. Thanks. – MathGod Dec 21 '14 at 20:33
  • @MathGod: We know a bunch about $k$. It can be written as $k_1k_2^2$ where $k_1$ is squarefree. It's either of the form $2m$ with $m$ odd, or can be written as $(\tfrac{k+1}{2})^2 - (\tfrac{k-1}{2})^2$ if odd or $(m+1)^2-(m-1)^2$ if even. etc. – Kieren MacMillan Dec 22 '14 at 02:58
  • As for your other question, feel free to email me if you want to chat about publication options. – Kieren MacMillan Dec 22 '14 at 03:00

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There is a well known elementary proof that the equation

$$ y^2 = x^3+7 $$

has no solution in integers $x,y$.

If $x$ is even, then $y^2 \equiv 3\pmod{4}$. Since this is not possible, $x$ must be odd. Hence $x^2-2x+4 \equiv 1+2x \equiv 3\pmod{4}$, and so there exists a prime $q$ such that $q \mid (x^2-2x+4)$ and $q \equiv 3\pmod{4}$. But then

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

implies $q \mid (y^2+1)$, which is impossible for primes of the form $4k+3$.

We have shown that the Mordell equation corresponding to $k=7$ has no integer solutions. $\blacksquare$

AT1089
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    How does this answer the original question? – Favst Jul 11 '20 at 17:18
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    This is the "elementary proof" for the case $k=-7$. See my comment below the OP, in which this was asked and answered. – Kieren MacMillan Jul 11 '20 at 17:25
  • Can somebody please tell me why this was downvoted? – AT1089 Jul 17 '20 at 04:07
  • @AT1089 it was probably downvoted because it does not answer the question. It does however give some insight about the case k = -7 but that was already mentioned in the comments below the OP. I upvoted btw. In my opinion it is neither good nor bad so a score of 0 seems reasonable. Less than zero is a punishment despite effort. – mick Aug 10 '20 at 19:19