# On the Diophantine equation $x^2 = y^p + 2^{r}z^p$ where $p\geq 7$ is an odd prime and $r \geq 2$

It is known that the only nonzero pairwise coprime integer solutions to the above Diophantine equation are for $r=3$, for which $(x, y, z) = (3,1,1)$ and $(-3,1, 1)$. (Cohen, Number Theory Volume 2: Analytic and Modern Tools pp 507-509, Theorem 15.3.4).

When one only assumes that $\gcd(x, y, z)=1$, we also have the solution $(x, y, z) = (3\cdot2^{\frac{r}{2}},2,1)$, for $r=p-3$. But are there any other nonzero integral solutions with none of $(\mid x \mid, \mid y \mid, \mid z \mid)$ being equal to $1$ ?

• You mention this about the tags each time you ask a question. But I see no reason why you shouldn't be able to add the tags yourself. – Lucia Dec 9 '15 at 22:29
• @Lucia, Well, each time i try to use the relevant tags, i get a notice that such and such a tag requires a certain minimum reputation (which i currently don't have since i'm still new here). I would indeed be grateful if you may share with me on how i can go about it. – user83236 Dec 9 '15 at 22:36
• I just checked it out from a different browser. I didn't encounter any problems in adding any tags to a fake question (while not logged into Mathoverflow). You could ask this in Meta, if indeed you're having some difficulties with the site. – Lucia Dec 9 '15 at 22:39
• @Tatenda: You can use any existing tag, also nt.number-theory. -- You only cannot create new tags, as this requires 300 points, cf. mathoverflow.net/help/privileges/create-tags. – Stefan Kohl Dec 9 '15 at 22:39

"Nous definissons de meme le fait pour $(a, b, c)$ de etre une solution propre non trivile de l'equation
Theoreme (a): Let $S_{0}(r,p)$ denote the set of all nonzero integral solutions to the Diophantine equation concerned. (this was the only part that i could translate into English). Si $r$ distinct de $1, 3, p-3$ et $p-1$, l'énsemble $S_{0} (r,p)$ est vide."
From my very limited understanding of French, i think its saying that $S_0(r, p)$ is empty if $r$ is not equal to $1,3, p-1$ or $p-3$.