System of quadratic Diophantine equations

Question

Is there a method for determining if a system of quadratic diophantine equations has any solutions?

My specific example (which comes from this question) is: $$\frac{4}{3}x^2 + \frac{4}{3}x + 1 = y^2$$ $$\frac{8}{3}x^2 + \frac{8}{3}x + 1 = z^2$$ I want to know if there are any positive integer triples $(x,y,z)$ which satisfy both equations.

Answer 1

We consider only the general question. It was proved by Matijasevich that there is no algorithm which, on input any Diophantine equation $P(x_1,x_2,\dots,x_m)=0$, where $P$ is a polynomial with integer coefficients, will determine whether the equation has an integer solution.

Using a little trick that goes back to Skolem, given any Diophantine equation $P(x_1,x_2,\dots,x_m)=0$, we can algorithmically produce a system $Q_i(y_1,y_2, \dots, y_n)=0$, $i=1,\dots, s$ of quadratic Diophantine equations such that the system has a solution in integers if and only if $P(x_1,x_2,\dots,x_m)=0$ has a solution in integers.

It follows that there is no algorithm which, given any system of quadratic Diophantine equations, will determine whether the system has a solution in integers.

Answer 2

Incomplete... First take $w = 2 x + 1$

From $z^2 - 2 y^2 = -1$ we get the sequence of $y$ values $$ y_0 = 1, \; \; y_1 = 5, \; \; y_{n+2} = 6 y_{n+1} - y_n, $$ or $$ 1,5,29,169,985,5741,33461, 195025 \ldots $$

From $w^2 - 3 y^2 = -2$ we get the sequence of $y$ values $$ Y_0 = 1, \; \; Y_1 = 3, \; \; Y_{n+2} = 4 Y_{n+1} - Y_n, $$ or $$ 1,3,11,41,153,571,2131, 7953, 29681, 110771 \ldots $$

The question can then be written about the two sequences. They grow at different rates, so the indices would be different anyway; but, are there indices $i,j$ with $$ y_i = Y_j? $$

At the least, the matter can be experimented with further this way: both sequences have explicit recipes involving square roots and exponents, same as the Fibonacci numbers and the golden ratio. The two roots of the characteristic polynomials are real, and after a relatively short while you can ignore the power of the smaller roots, and compare $y_i$ and $Y_j$ using logarithms to find good pairs $i,j.$ Finally, as mentioned, I imagine this can be finished with some algebraic number theory. But you could also finish it by showing $\log y_m$ and $\log Y_n$ never get quite close enough.

For what it may be worth, $$ y_n = \left( \frac{\sqrt 2 + 1}{\sqrt 8} \right) \left( 3 + \sqrt 8 \right)^n + \left( \frac{\sqrt 2 - 1}{\sqrt 8} \right) \left( 3 - \sqrt 8 \right)^n $$ and $$ Y_n = \left( \frac{\sqrt 3 + 1}{\sqrt {12}} \right) \left( 2 + \sqrt 3 \right)^n + \left( \frac{\sqrt 3 - 1}{\sqrt {12}} \right) \left( 2 - \sqrt 3 \right)^n $$

Answer 3

Answered at the source question. Briefly: There are no such $(x,y,z)$, because then we'd have a nonconstant arithmetic progression of four squares: $$ 1 = 1^2, \ \frac43(x^2+x)+1 = y^2, \ \frac83(x^2+x)+1 = z^2, \ 4(x^2+x)+1 = (2x+1)^2. $$ The impossibility of such a progression is a theorem of Euler (1780, answering a question "first raised by Fermat in 1640" according to Keith Conrad's exposition). This even proves that there are no rational solutions $(x,y,z)$ other than the obvious ones with $x=0$ or $x=-1$.