Pair Sum Equals Single Sum

We Can Rewrite As

$(a-1)(b-1)+(c-1)(d-1)=2$

Now There Are Not Many Possibilities to Consider ! If the First Product Is $0$ , The Second Must Be $2$ , and If the First Product Is $1$ , So is The Second.

If $a=1$ , Then We Need to Have $(c-1)(d-1)=2$ . Since $1≤c≤d$ , This Forces $c=2, d=3$ . And $b$ Can Be $1$ or $2$ , Giving the Solutions $(1,1,2,3)$ and $(1,2,2,3)$ .

If $a>1$ , We Need $a=2$ , Else the Left Hand Side Is Too Big. That Forces $b=c=d=2$ , Giving the Third Solution $(2,2,2,2)$ .

Sets of Solutions : ( $1,1,2,3),(1,2,2,3), (2,2,2,2)$ .