Find quadruples of positive non-zero integers $(a, b, c, d)$, such that all pairwise products of the four numbers plus one are square numbers. Consider $1 \leq a < b < c < d < N$. How many quadruples exist if $N = 100000$?
Using a brute force approach, for $N = 1000$, I find three solutions: $(1, 3, 8, 120)$, $(1, 8, 15, 528)$ and $(2, 4, 12, 420)$. For the third quadruple:
- $2 \cdot 4 + 1 = 3 \cdot 3$
- $2 \cdot 12 + 1 = 5 \cdot 5$
- $2 \cdot 420 + 1 = 29 \cdot 29$
- $4 \cdot 12 + 1 = 7 \cdot 7$
- $4 \cdot 420 + 1 = 41 \cdot 41$
- $12 \cdot 420 + 1 = 71 \cdot 71$
I assume there is no simple answer for this, but is it solvable at all? Can it be connected with existing problems with known/unknown solutions?
Best Answer
Greg Martin shows that the number of such quadruples up to $N$ is asymptotic to $CN^{1/3}\log N$, where $C={2^{4/3}\over3\Gamma(2/3)^3}$, which is about $0.33828$.