Jae

Given that $\begin{cases} ab+c+d = 15 & ...(1) \\ bc+d+a = 19 & ...(2) \\ cd + a + b = 25 &...(3) \\ da + b + c = 17 & ...(4) \end{cases}$

$\begin{cases} (2)-(1): & (b-1)(c-a) = 4 \\ (3)-(4): & (d-1)(c-a) = 8 \end{cases} \implies \dfrac {d-1}{b-1} = 2 \implies d = 2b-1$

$\begin{cases} (3)-(2): & (c-1)(d-b) = 6 \\ (4)-(1): & (a-1)(d-b) = 2 \end{cases} \implies \dfrac {c-1}{a-1} = 3 \implies c = 3a-2$

From equation 2:

$\begin{aligned} bc+d+a & = 19 \\ b(3a-2)+2b-1+a & = 19 \\ 3ab + a & = 20 \\ \implies b & = \frac {20-a}{3a} &...(2a) \end{aligned}$

From equation 4:

$\begin{aligned} da + b + c & = 17 \\ (2b-1)a + b + 3a-2 & = 17 \\ 2ab + 2a + b & = 19 \\ (2a+1){\color{#3D99F6}b} + 2a & = 19 & \small \color{#3D99F6} (2): \ b = \frac {20-a}{3a} \\ (2a+1)\times \frac {20-a}{3a} + 2a & = 19 \\ (2a+1)(20-a) + 6a^2 & = 57a \\ -2a^2 + 39a + 20 + 6a^2 & = 57a \\ 4a^2 - 18a + 20 & = 0 \\ 2a^2 - 9a + 10 & = 0 \\ (2a-5)(a-2) & = 0 \\ \implies a & = 2 & \small \color{#3D99F6} \text{Since } a \in \mathbb N \\ b & = \frac {20-a}{3a} = 3 \\ c & = 3a-2 = 4 \\ d & = 2b-1 = 5 \end{aligned}$

Therefore, $a$ , $b$ , $c$ and $d$ are 2, 3, 4 and 5 respectively.