<s>[INST] $a+b=c \times d$ and $a\times b = c + d$ There is a 'nice' relationship between the integers (1,5) and (2,3) as 
$$1+5=2 \times 3;$$
$$1\times 5 = 2 + 3.$$
So I tried to find all positive integers pairs $(a, b)$ and $(c, d)$ such that $$a+b=c \times d;$$ 
$$a\times b = c + d.$$
To find this, $a, b, c, d$ must satisfy $$(a+1)(b+1)=(c+1)(d+1).$$ However, this condition is only necessary but not sufficient.
Any idea?
 [/INST] The following approach is based on the idea that, for positive integers $x,y$,  the product $xy$ typically exceeds the sum $x+y$. We can apply this to show that in the problem of the question, either $a=b=c=d=2$ or else at least one of $a,b,c,d$ is 1; in this case it's easy to arrive at the only remaining solution $1,5;2,3$.
Suppose for two positive integers $x$ and $y$, that $x,y \ge k$ for some fixed $k \ge 2$. Then from $(x-k)(y-k) \ge 0$ we have, after expanding and rearranging, that $xy \ge k(x+y-k)$. 
Now assume that each of $a,b,c,d$ is at least $k$.  Of course we also use the assumptions $ab=c+d,cd=a+b$ of the question. We then have $ab \ge k(a+b-k)=k(cd-k)$, and also $cd \ge k(c+d-k)=k(ab-k)$. Putting these together we have 
$ ab \ge k(k(ab-k)-k)=k^2ab-(k^3+k^2)$. 
This inequality simplified is
$$ab \le \frac{k^2}{k-1}.$$
Now if $k=2$ here we obtain $ab \le 4$ which with the assumptions  $a,b \ge 2$ leads to $a=b=2$, and (by symmetry or by the initial relations) also $c=d=2$.
If $k=3$ then we obtain $ab \le 9/2 = 4.5$, but this cannot hold since we're assuming $a,b \ge 3$ so that in fact $ab \ge 9$.
So by considering how products are typically larger than sums, we have shown that, except for the solution $a=b=c=d=2$, one of the values $a,b,c,d$ must be 1. Putting say $a=1$ into the two equations, one easily gets $a=1,b=5$ and that  $c,d$ are $2,3$ insome order.
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