If $p \geq 5$ is a prime, are there any integers $x, y, z > p$ such that $(x, y) = 1$ and $$x^{p} - 4y^{p} = z^{2}$$
Asked
Active
Viewed 72 times
2
-
There's $5^3-4\cdot 1^3=11^2$ ... – Hagen von Eitzen Aug 19 '14 at 13:32
-
Hi thanks. It reminds me to put the condition $x, y, z > p.$ – Yes Aug 19 '14 at 13:52
-
See http://mathoverflow.net/questions/178906/the-diophantine-equation-xp-4yp-z2. – Dietrich Burde Aug 20 '14 at 08:06
-
@Dietrich Burde Yes, thank you. It is I who asked the same question in MatheOverflow. – Yes Aug 20 '14 at 08:08
-
All right, very good. So there is a literature on this problem. I was looking for this article of Bennett and Skinner, but could not find it. Is there a reference for $p=3$ somewhere ? – Dietrich Burde Aug 20 '14 at 08:14
-
Cohen has treated the equation $x^3-2y^3=z^2$ in his book on Diophantine equations. – Dietrich Burde Aug 20 '14 at 08:28
1 Answers
2
This equation is a special case of the generalized Fermat equation $$ Ax^p+By^q=Cz^r $$ for $A=C=1$, $B=-4$ and $p=q$, $r=2$. We have $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1$ in our case $(p,q,r)=(p,p,2)$ for $p>3$, so that we are in the hyperbolic case. Hence for $p>3$ there are at most finitely many coprime solutions $(x,y,z)$ by the Darmon-Granville theorem. And probably there are very few such solutions - the abc-conjecture implies that there are at most $2$ solutions once $n>n_0$, independent of $A,B,C$. The equation is not hyperbolic for $p=3$, but the argument with the abc-conjecture still applies. There is a large literature on the generalized Fermat equation, which will be helpful to study this case (the experts might know more).
Edit: A reference was given afterwards here, and Gerry Myerson found $78^3-4\cdot 29^3=614^2$ and $93^3-4\cdot 53^3=457^2$for $p=3$.
Dietrich Burde
- 130,978
-
(Maple and) I found 8 examples of $x^3-4y^3=z^2$ with $\gcd(x,y)=1$ and $x<1000$. – Gerry Myerson Aug 20 '14 at 13:06