Find all quadruples $(a, b, c, d)$ of non-negative integers such that $a b = 2(1 + c d)$ and there exists a non-degenerate triangle with sides of length $a - c$, $b - d$, and $c + d$.

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It holds that $a>c$ and $b>d$ because $a-c$ and $b-d$ must be sides of a triangle. Therefore, $a \geq c+1$ and $b \geq d+1$, since they are integers. We now distinguish two cases: $a>2 c$ and $a \leq 2 c$.

Suppose that $a>2 c$ holds. Then $a b>2 b c \geq 2 c \cdot(d+1)=2 c d+2 c$. On the other hand, $a b=2+2 c d$, so $2 cd$, so $b \geq 2$. From $a b=2$ it follows that $a=1, b=2$. Therefore, $d=1$ and the triangle has sides of length 1, 1, and 1. Such a triangle indeed exists. Thus, the quadruple $(1,2,0,1)$ is indeed a solution.

Now suppose that $a \leq 2 c$ holds. The triangle inequality states that $(a-c)+(b-d)>c+d$, so $a+b>2(c+d)$. Since $a \leq 2 c$, it follows that $b>2 d$. We also knew that $a \geq c+1$, so $a b>(c+1) \cdot 2 d=2 c d+2 d$. On the other hand, $a b=2+2 c d$, so $2 d<2$. This means that $d=0$. Analogous to the case $c=0$, this results in the only solution being the quadruple $(2,1,1,0)$.

The only solutions are thus $(1,2,0,1)$ and $(2,1,1,0)$.