$a,b,c,d$ are positive integers such that $ad=bc$. Prove that $n=a+b+c+d$ cannot be prime

I'm not sure how to end your solution, but I have another one if you are interested:

Suppose $a+b+c+d=p\in \mathbb{P}$. Since $d=p-a-b-c$ we get $$a(p-a-b-c)=bc$$ so $$ ap = (a+b)(a+c) \implies p\mid a+b \;\;\;\;{\rm or }\;\;\;\;p\mid a+c$$

A contradiction (since $p>a+b$ and $p>a+c$).

Since $d = bc/a$ is an integer, we can factor $d = (b/x)(c/y)$ for some positive integers $x,y$ such that $b/x$ and $c/y$ are integers and $xy=a$. (This is intuitively clear and can be made rigorous by playing around with prime factorizations of $a,b,c$.)

Then $$ a + b + c + d = xy + b + c + (b/x)(c/y) = (x + c/y)(y + b/x) $$ which is always a nontrivial factorization, so $a+b+c+d$ cannot be prime.

Here's one way to do it. Maybe not the best.

Let $w = \gcd(a,b)$ so that $a = a'w$ and $b=b'w$ so we have

Let $v = \gcd(c,d)$ so that $c =c'v$ and $d=d'v$ so we have

$ad = bc$

$a'd'wv = b'c'wv$ so $a'd' =b'c'$. So we have $b'|a'd'$ but as $a',b'$ are relatively prime $b'|d'$. And we also have $d'|b'c'$ but as $d',c'$ are relatively prime we have $d'|b'$.

So $b'|d'$ and $b'|d'$ and (these are positive values) so $d' = b'$. And so

$a'd' = d'c'$ so $c'= a'$.

And... that's it really.

$a + b + c + d= a'w + b'w + c'v + d'v = a'w+ d'w + a'v + d'v= (a'+d')(v + w)$

As we assumed $a,b,c,d> 0$ then $a',d',v,w \ge 1$ and $(a'+d'), (v+w)\ge 2$. Thus it's a composite number.

All the posted answers are more or less the same. It's just a matter of which tack seems clearer and straight forward. Mine isn't the fastest but it "falls out" pretty nicely.

It also gives us a useful lemma:

If $ab = cd$ then $\frac a{\gcd(a,b)} = \frac c{\gcd(c,d)}$ and $\frac b{\gcd(a,b)} = \frac d{\gcd(c,d)}$ .