Distinct positive integers $x_1,x_2,x_3,x_4$ that satisfy $4x_1^2=x_2^2+2x_3^2+x_4^2.$

Question

So I want to find a solution to the Diophantine equation $4x_1^2=x_2^2+2x_3^2+x_4^2,$ such that $x_1,x_2,x_3$ and $x_4$ are distinct positive integers which also satisfy the inequalities $3x_1^2-x_3^2-x_4^2>0$ and $3x_1^2-x_2^2-x_3^2>0.$ I don't know how to go about doing this though, or if it's even possible. Ideas would be appreciated.

Answer 1

This is not a complete solution.

$$4x_1^2 = x_2^2 + 2x_3^2 + x_4^2\tag{1}$$

First, we get the parametric solution of equation $(1).$
Next, we check the inequalities $3x_1^2-x_3^2-x_4^2>0$ and $3x_1^2-x_2^2-x_3^2>0.$

Substitute $x_1=p+a, x_2=p+b, x_3=p+c, x_4=p+d$ to equation $(1)$, we get

$$p = \frac{-1}{2}\frac{4a^2-d^2-2c^2-b^2}{-b+4a-d-2c}$$

Hence we get the parametric solution of equation $(1).$
$x_1 = 4a^2+d^2+2c^2+b^2-2ab-2ad-4ac$
$x_2 = 4a^2-d^2-2c^2+b^2-8ab+2bd+4bc$
$x_3 = 4a^2-d^2+2c^2-b^2+2bc-8ac+2dc$
$x_4 = 4a^2+d^2-2c^2-b^2+2bd-8ad+4dc$

The solutions that satisfy the two inequalities are as follows.

           (a,b,c,d) (x1,x2,x3,x4)
       (1,3,4,2),( 23, 13, 31,  5)
       (1,2,5,3),( 37, 15, 51,  7)
       (1,6,2,5),( 43, 67, 21, 45)
       (1,4,5,3),( 45, 33, 59,  7)
       (1,6,3,5),( 49, 81,  3, 55)
       (1,2,6,3),( 55, 29, 75,  3)
       (1,6,4,5),( 59, 91, 31, 61)
       (1,7,2,6),( 63, 93, 37, 67)
       (1,5,6,2),( 67, 53, 83, 37)
       (1,5,6,4),( 75, 61, 95, 27)
       (1,2,7,3),( 77, 47,103, 17)
       (1,7,4,6),( 79,125, 23, 91)
       (1,7,6,2),( 87,117, 83, 53)
       (1,2,7,5),( 89, 55,115, 47)
       (1,7,5,6),( 93,135, 59, 97)
       (1,8,3,7),( 93,145, 25,111)
       (1,5,7,4),( 97, 55,131, 17)
       (1,6,7,3),(101, 89,127, 25)
       (1,8,4,7),(103,163, 11,125)
       (1,2,8,3),(103, 69,135, 35)
       (1,3,8,4),(111, 35,155,  3)
       (1,6,7,5),(113, 97,139, 55)
       (1,9,2,8),(115,157, 81,123)
       (1,2,8,5),(115, 77,151, 37)
       (1,9,3,8),(121,183, 45,145)
       (1,8,7,3),(125,169,127, 41)
       (1,6,8,3),(127, 83,167, 43)
       (1,9,4,8),(131,205,  5,163)
       (1,8,6,7),(135,187, 95,141)
       (1,9,7,2),(137,199,115, 95)
       (1,8,7,5),(137,185,139, 47)
       (1,6,8,5),(139, 91,183, 45)
       (1,7,8,4),(143,133,179,  5)
       (1,2,9,5),(145,103,191, 23)
       (1,9,5,8),(145,223, 39,177)
       (1,5,9,4),(153, 31,215, 15)
       (1,6,9,3),(157, 73,211, 65)
       (1,7,8,6),(159,141,191, 91)
       (1,9,7,6),(161,239,139, 89)
       (1,9,8,2),(163,205,159,117)
       (1,9,6,8),(163,237, 87,187)
       (1,2,9,7),(165,119,203,111)
       (1,6,9,5),(169, 81,231, 31)
       (1,9,8,4),(171,229,179, 21)
       (1,7,9,4),(173,127,227, 23)
       (1,8,9,3),(181,169,219, 81)
       (1,9,7,8),(185,247,139,193)
       (1,9,8,6),(187,245,191, 83)
       (1,8,9,5),(193,185,239, 23)
       (1,8,9,7),(213,193,251,135)

Answer 2

I don't know of systematic ways of solving these sorts of problems, but one thing that can give you information about what possible solutions there could be is looking at the equation with respect to different moduli. For instance, in $\mathbb Z_4$, $0^2=2^2=0$ and $1^2=(-1)^2=1$, so $4x_1^2$ must be $0$, $x_2^2$ $x_4^2$ can be $0$ or $1$, and $2x_3^2$ can be $0$ or $2$. If $2x_3^2=0$, then $x_2^2$ and $x_4^4$ have to both be $0$, and if $2x_3^2=2$, then $x_2^2$ and $x_4^4$ have to both be $1$. From this we can conclude that the parity of $x_2$, $x_3$, and $x_4$ must be the same.

Answer 3

There are several papers giving complete solutions to 2.m.n Diophantine equation(s) — yours is the 2.1.4 equation with a few special constraints. I would first suggest looking at Bradley’s paper, and then Barnett et al.