SAUDITYA wrote: $X = \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}$ ; $Y = \frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}$. See that $X+Y =4$ .So $X \in \mathbb{Z} \iff Y \in \mathbb{Z}$. By C-S, $X,Y>1$ Why is $X>1$?

Because $$\sum_{cyc}\frac{a}{a+b}>\sum_{cyc}\frac{a}{a+b+c+d}=1.$$

Let $$A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a} \hspace{0.43cm}\text{and}\hspace{0.43cm} B=\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a},$$then $A+B=4$. Therefore, $A<4$. On the other hand, let us note that, $$A\cdot (a+b+c+d)=\sum_{cyc}\left(a+\frac{a(c+d)}{a+b}\right)>a+b+c+d.$$Therefore, $1<A<4$. Similarly, $1<B<4$. This means that $A\in\left\{2, 3\right\}$, but if $A=3$, then $B=1$, which is absurd. Therefore, $A=2$. Suppose for the absurdity that $a+b+c+d=p$, where $p$ is prime, and let $M=(a+b)(b+c)(c+d)(d+a)$, then $$c+d\equiv -(a+b)\pmod p,\hspace{0.52cm} a+d\equiv -(b+c)\pmod p \hspace{0.52cm}\text{and}\hspace{0.52cm} M\equiv (a+b)^2(b+c)^2\pmod p.$$Furthermore, $$\begin{matrix}\vspace{0.43cm} \dfrac{M}{a+b}&\equiv&(a+b)(b+c)^2\pmod p \\ \vspace{0.43cm} \dfrac{M}{b+c}&\equiv&(a+b)^2(b+c)\pmod p \\ \vspace{0.43cm} \dfrac{M}{c+d}&\equiv& -(a+b)(b+c)^2\pmod p \\ \vspace{0.43cm} \dfrac{M}{d+a}&\equiv& -(a+b)^2(b+c)\pmod p \end{matrix}$$Since $$\frac{aM}{a+b}+\frac{bM}{b+c}+\frac{cM}{c+d}+\frac{dM}{d+a}=2M,$$and taking into account that $a+b$ and $b+c$ are coprimes with $p$, then in module $p$ we have to $$a(b+c)+b(a+b)-c(b+c)-d(a+b)\equiv 2(a+b)(b+c)\pmod p.$$Reducing this last congruence, we have to $$(a+b)^2\equiv (b+c)^2\pmod p,$$from which we deduce that $a+b\equiv b+c\pmod p$ or $a+b\equiv -(b+c)\pmod p$. Therefore, $a\equiv c\pmod p$ or $b\equiv d\pmod p$. Since $a$, $b$, $c$, $d<p$, then $a=c$ or $b=d$, which is absurd. Therefore, $a+b+c+d$ is not prime.

Let $X = \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}, Y= \frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}$. Note that: $X+Y=4$ and $X, Y$ are integers. Notice that: $$X = \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a} > \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d} = 1.$$Thus, $X=Y=2$. Then, we have: $$0 = X-Y = \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-d}{c+d}+\frac{d-a}{d+a} = \frac{(a-b)(c+d)+(a+b)(c-d}{(a+b)(c+d)} + \frac{(a+d)(b-c)+(b+c)(d-a}{(a+d)(b+c)}$$$$=2(ac-bd)\frac{(a-c)(b-d)}{(a+b)(b+c)(c+d)(d+a)}.$$ Thus, we have $ac=bd$. Note that, there exists some positive integers $p, q, r, s$ such that: $a=pq, b=qr, c=rs, d=sp$. Therefore $a+b+c+d=(p+r)(q+s)$ is not a prime number.

Brilliant problem. My solution is basically the same as above ones. We let $X = \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}, Y= \frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}$. Note that $X+Y=4$ and $X, Y$ are both positive integers (since $X$ is given to be one). Further, \[X = \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a} > \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d} = 1\]So we must have $X=Y=2$. Now, \begin{align*} 0 &= X-Y\\ &= \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-d}{c+d}+\frac{d-a}{d+a} \\ &= \frac{(a-b)(c+d)+(a+b)(c-d}{(a+b)(c+d)} + \frac{(a+d)(b-c)+(b+c)(d-a}{(a+d)(b+c)}\\ & =2(ac-bd)\frac{(a-c)(b-d)}{(a+b)(b+c)(c+d)(d+a)} \end{align*}Now since $a\ne c$ and $b \ne d$ this implies that $ac=bd$. By the Four Number Lemma, this implies that there exists positive integers $p,q,r,s$ such that $a=pq$ , $b=qr$ , $c=rs$ and $d=sp$. But then, note that \[a+b+c+d=pq+qr+rs+sp = (p+r)(q+s)\]which is most clearly non-prime.