<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] There is no other solution for positive integers because if we consider the pairs $(n, x)$ and $(m, y)$ such that $$n+x=m \times y;$$ 
$$n\times x = m + y$$
We have $x=my-n$ and $m+y=n(my-n)=nmy-n^2$ and $y-nmy=-n^2-m$ and finally $$y=\frac{n^2+m}{nm-1}$$ 
If we supose that $n$ is a positive integer fixed and $m$ variable this is a hyperbola. 
For the case n=1 we have $y(2)=3$ are the only integers points and in $x=my-n=2\times 3-1=5$ we find $(1,5)$ and $(2,3)$ as we are expecting. 
If we consider the case $n=2$ then  $y(2)=2$ and in $x=my-n=2\times 2-2=2$ we find $(2,2)$ and $(2,2)$ trivially satisfies the condition. 
For $n=3$ we have $y(5)=1$ and in $x=my-n=5\times 1-3=2$ we find $(3,2)$ and $(5,1)$.
But for $n>3$ don't exist intenger $m$ such that $y(m)$ is integer, then there is not more solutions for the condition.
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