Let prime number p be a factor of positive integer m. The integer m has an expression as the difference between two perfect squares s and t, with s greater than t, such that either s+t or s-t equals p. When m=p×1, this will be the only such s and t, and s=t+1.

Define the ceiling square of a positive number as the smallest perfect square greater than or equal to the number. The only primes for which s in the above relationship is the square root of p’s ceiling square are 3 and 5. (*N.B.: These are the only numbers with a cyan background in the “integer triangle” depicted in my previous post.)*

EDITED TO ADD: Also, I have a follow-up questions to these observations, for anyone who is interested: Does this transform the factorization of an integer m=pq, p<q, into the process of finding s and t such that s^{2}-t^{2}=m and s-t=p? I think it does, but I also expect it doesn’t simplify anything. I have shared this question on Twitter already, tagging the American Mathematical Society account there. My hopeful (perhaps unreasonably so) heart says it’s worth a shot.

## Leave a Reply Cancel reply