Numbers with Four Close Factorizations
Disciplines
Number Theory
Abstract (300 words maximum)
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.
Use of AI Disclaimer
no
Academic department under which the project should be listed
CSM – Mathematics
Primary Investigator (PI) Name
Dr. Tsz Chan
Numbers with Four Close Factorizations
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.