Approach 1: Adding up the equations, $30 (a + b + c + d) = a^2 + b^2 + c ^2 + d^2 \geq \frac{ (a+b+c+d) ^2 } { 4} $.
Since $ a+b + c + d > 0$, we can divide by it and conclude that hence $ 120 \geq a+b+c+d$.
WLOG, let $ a= \min (a, b, c, d)$.
Then $ a^2 = 9b + 10c + 11d \geq 9a + 10 a + 11a = 30a$, so $ a \geq 30$.
The only possible tuple which satisfies both conditions is $(30,30,30,30)$, which we can easily verify is a solution.
Approach 2: (Fill in the gaps yourself) Similar to the 2nd step from above, show that
$$ 30 \leq \min (a, b, c, d) \leq \max (a, b, c, d ) \leq 30. $$
Note:
- I believe (no proof yet) that $ (0,0,0,0)$ is the only other real solution.
- Wolfram says that there are other complex solutions.
You can find the minimum such $x$ via integer linear programming. For example, here is SAS code to do that:
data indata;
input i d r m;
datalines;
1 100 4 8
2 101 2 8
3 105 3 8
4 106 7 8
;
proc optmodel;
/* declare parameters and read data */
set OBS;
num d {OBS};
num r {OBS};
num m {OBS};
read data indata into OBS=[i] d r m;
/* declare decision variables */
var X >= 1 integer;
var Q {OBS} >= 0 integer;
/* declare objective */
min Z = X;
/* declare constraints */
con C1 {i in OBS}:
X >= d[i] * (r[i] + m[i] * Q[i]);
con C2 {i in OBS}:
X <= d[i] * (r[i] + m[i] * Q[i] + 1) - 1;
/* call mixed integer linear programming solver */
solve;
quit;
The resulting log is as follows:
NOTE: Problem generation will use 4 threads.
NOTE: The problem has 5 variables (0 free, 0 fixed).
NOTE: The problem has 0 binary and 5 integer variables.
NOTE: The problem has 8 linear constraints (4 LE, 0 EQ, 4 GE, 0 range).
NOTE: The problem has 16 linear constraint coefficients.
NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range).
NOTE: The OPTMODEL presolver is disabled for linear problems.
NOTE: The initial MILP heuristics are applied.
NOTE: The MILP presolver value AUTOMATIC is applied.
NOTE: The MILP presolver removed 0 variables and 4 constraints.
NOTE: The MILP presolver removed 8 constraint coefficients.
NOTE: The MILP presolver modified 0 constraint coefficients.
NOTE: The presolved problem has 5 variables, 4 constraints, and 8 constraint coefficients.
NOTE: The MILP solver is called.
NOTE: The parallel Branch and Cut algorithm is used.
NOTE: The Branch and Cut algorithm is using up to 4 threads.
Node Active Sols BestInteger BestBound Gap Time
0 1 0 . 742.0000000 . 0
0 1 0 . 20246.0000000 . 0
0 1 0 . 35595.0000000 . 0
0 1 0 . 42000.0000000 . 0
0 1 0 . 70278.0000000 . 0
0 1 0 . 128674 . 0
0 1 0 . 158867 . 0
0 1 0 . 173922 . 0
0 1 0 . 186042 . 0
0 1 0 . 192506 . 0
0 1 0 . 258800 . 0
NOTE: The MILP solver added 1 cuts with 2 cut coefficients at the root.
62 0 1 589254 589254 0.00% 0
NOTE: Optimal.
NOTE: Objective = 589254.
Best Answer
ADDED: pretty much everything needed for this answer is in this chapter of BUELL
The method suggested in comments gives first $a = d + 2c$ and then $d = 5b-7c.$ Plugging those into $ad-bc=1$ gives $$ 25 b^2 - 61bc + 35 c^2 = 1. $$ If we had $b=c = 1$ the quadratic form would evaluate to $-1.$ However, $1$ itself is impossible. The printout below shows the Gauss-Lagrange method of "reduced" indefinite binary quadratic forms. It is a theorem of Lagrange that all small numbers (below $\frac{1}{2} \sqrt {221}$ in absolute value) that are primitively integrally represented by $\langle 25 -61, 35 \rangle$ must appear as first or last coefficients of a form in the chain of reduced forms equivalent to the original; however
This is equivalent to the fact that $x^2 - 221 y^2 \neq -1$ for integers $x,y,$ proof by continued fractions, really: