Consider n = 99, 990, 000, a number that has two close factorizations: 10, 000 · 9, 999 and 11,000 · 9, 090. Generalizing this to k close factorizations, we have n = AB = (A + a_1)(B − b_1) = (A + a_2)(B − b_2) = · · · = (A + a_{k−1})(B − b_{k−1}) where 1 ≤ B ≤ A as well as 1 ≤ a_1 < a_2 <· · · < a_{k−1} ≤ C and 1 ≤ b_1 < b_2 < · · · < b_{k−1} ≤ C. Here, C is our closeness measure. Our faculty mentor, Dr. Tsz Chan, previously studied numbers with three close factorizations and identified an optimal ratio R_3 = A / C^3 ≤ 0.25. The scope of our project this summer was to expand on his work and study numbers with four close factorizations. The optimal closeness ratio here was calculated to be R_4 = A / C^3 ≤ (6+√ 6) / 9(2+√ 6)^2 = 0.04742…. Arriving at this ratio involved proving identities and inequalities, deducing intermediate lemmas, transforming our original question into a generalized Pell-type equation, determining which numbers should be entered into that equation, and eliminating cases that we knew would yield no solutions.