Second degree Diophantine equations I found a question whether there are general methods to solve second degree Diophantine equations. I was unable to find an answer so is this known? In particular, the original writer wants to know whether one can find all integers satisfying $x^2 + x = y^2 + y + z^2 + z$.
 A: About algorithm: There is an algorithm that will determine, given any quadratic $Q(x_1,\dots,x_n)$ as input, whether or not the Diophantine equation $Q(x_1,\dots,x_n)=0$ has a solution. This is something that I (and others) observed quite a long time ago. I have no knowledge about a nice algorithm.
Set one machine $M_1$ to search systematically for solutions.  Another machine $M_2$ simultaneously checks whether there is a real solution (easy) and then checks systematically for every modulus $m$ whether there is a solution modulo $m$. 
By the Hasse Principle (which in this case is a theorem), if our equation has "local" solutions (real and modulo $m$ for every $m$) then it has an integer solution. So either $M_1$ will bump into a solution or $M_2$ will find a local obstruction to a solution. Thus the algorithm terminates.
The corresponding question for cubics is unsolved. The same question for quartics (in arbitrarily many variables) is equivalent to the general problem of testing a Diophantine equation for solvability, so is recursively unsolvable.
Added: I think that the details are written out in the book Logical Number Theory I by Craig Smorynski. Very nice book, by the way.  
A: There is an algorithm for deciding whether a single degree 2 multivariable polynomial equation has a solution in integers, due to Siegel, Zur Theorie der quadratischen Formen, Nachr. Akad. Wiss. Göttingen Math.-Phys. Kl. II 1972, 21-46.  See also Grunewald and Sigel, On the integer solutions of quadratic equations, J. Reine Angew. Math. 569 (2004), 13-45.  
But the Hasse principle by itself does not give an algorithm: it holds only for rational solutions, not for integer solutions.
A: Rewrite this equation a little differently.   $$X (X +a)+Y (Y +a)=Z (Z +a)$$   
Formulas for the solution can then be written, $p,k$ - where are integers and sets us.   
$$X =pk$$   $$Y =\frac{(p^2 −1)k}{2} +\frac{(p−1)a}{2}$$  $$Z =\frac{( p^2 +1)k}{2} +\frac{(p−1)a}{2}$$   
If we use the solutions of Pell's equation $p^2 −2 s^2 =1$  
Then the solution can be written:   $$X =2(s+p)sL+as(2s+p)$$  $$Y =(2s+p)pL+as(2s+p)$$  $$Z =(2 s^2 +2ps+ p^2 )L+2as(s+p)$$   
And more.   $$X =2s(s−p)L+ap(s−p)$$  $$Y =(p−2s)pL+ap(s−p)$$  $$Z =(2 s^2 −2ps+ p^2 )L+ap(2s−p)$$   
$L$ - given by us and can be any integer.
