Is it possible to find a number $N$ such that it has two factor pairs {a,b}, {c,d} such that (a+1)(b+1)=(c+1)(d+1)
My intuition tells me this is impossible in the case where a,b,c,d are positive integers. However I’m curious if this is still true for negative $N$ too. For negative $N$, assume $a$ and $c$ are negative.
Let $a, b, c, d, N \in \mathbb{Z}\setminus\{0\}$ such that $N = ab = cd$. Suppose that $(a+1)(b+1) = (c+1)(d+1)$.
Expanding $(a+1)(b+1) = (c+1)(d+1)$ we get that $a+b = c + d$.
Now substituting $b = \frac{N}{a}$ and $d = \frac{N}{c}$ we get $a+\frac{N}{a} = c + \frac{N}{c}$.
Rearranging we have $(a-c)(ac-N)=0$ and using that $N =ab$ we get $(a-c)(c-b) = 0$.
Hence the only solution is $\{a, b\} = \{c, d\}$.