Task 1. 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 $c + d$.
Solution I. We have $a>c$ 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: $a>2c$ and $a \leq 2c$.

Suppose that $a>2c$ holds. Then $ab > 2bc \geq 2c \cdot (d+1) = 2cd + 2c$. On the other hand, $ab = 2 + 2cd$, so $2cd$, so $b \geq 2$. From $ab = 2$ 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 $a \leq 2c$ holds. The triangle inequality states that $(a-c) + (b-d) > c + d$, so $a + b > 2(c + d)$. Since $a \leq 2c$, it follows that $b > 2d$. We also knew that $a \geq c+1$, so $ab > (c+1) \cdot 2d = 2cd + 2d$. On the other hand, $ab = 2 + 2cd$, so $2d < 2$. This means that $d=0$. Analogous to the case $c=0$, this yields the only solution as the quadruple $(2,1,1,0)$.

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