# Tag Info

Accepted

### Estimating the size of solutions of a diophantine equation

This problem turned out to be much more interesting than I originally thought. Let me give my solution, which seems to be slightly different from (but essentially the same as) the solution in the ...
Accepted

Accepted

### The "stubborn" solutions to sums of three cubes

This lim sup indeed goes to $\infty$. We can prove this using exactly the strategy Lucia suggested. We will count the number of $x,y,z$ in a box with $x^3+y^3+z^3$ not a cubic residue modulo $p$ for a ...
Accepted

### Why is this "the first elliptic curve in nature"?

I actually only wrote the part that says that this curve is a model for $X_1(11)$, not the first part, which I think was written by John Cremona. It is standard to order elliptic curves by conductor (...
Accepted

### Diophantine equation $3^n-1=2x^2$

This problem happens to have appeared on the Polish Mathematical Olympiad camp in 2015. Here is the official solution of the problem: (I use $m$ in place of $x$ because this is how the problem was ...

### Can $x^4+y^4+1$ be a perfect power?

To answer question 2: $$346^4+36788^4+1=1831575032204939793=3^3\cdot19^3\cdot179^2\cdot17569^2.$$

### Why does representing functors help solving Diophantine equations?

E.g. let $f(x,y, z)=0$ be a smooth projective plane curve with $f$ a rational polynomial of degree $\ge 4$. Then Mordell conjectured, and Faltings proved, that this has only finitely many rational ...

### Fermat's last theorem over larger fields

This is not quite an answer, but not quite a comment either. We can at least show that $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$ is infinite (where $X$ is the quintic Fermat curve). There are in ...

### Does the equation $(xy+1)(xy+x+2)=n^2$ have a positive integer solution?

It looks that Vieta jumping helps. For fixed positive integer $y$ choose a minimal positive integer $x$ for which $(xy+1)(xy+x+2)$ is a perfect square. Denote $4(xy+1)(xy+x+2)=4n^2=(2xy+x+3-z)^2$ ...
The perfect cuboid problem: We do not know if there is a common integer solution to $$a^2+b^2+c^2=d^2$$ $$a^2+b^2=e^2$$ $$a^2+c^2=f^2$$ $$b^2+c^2=g^2$$ with $a,b,c \ge 1$. The last condition can ...
There are bivariate coprime polynomial parametrizations: https://sites.google.com/site/tpiezas/010 $$(a^4-2ab^3)^3 + (a^3 b+b^4)^3 + (2a^3 b-b^4)^3 = (a^4+a b^3)^3$$ Added If you drop the positivity ...