Binomial-sum primality criterion for numbers of the form 4m plus or minus 1

Let mm be a positive integer. The binomial-sum primality conjecture. If m30m\neq30 and

k=04m2(2kk)mk4m1(mod(4m1)2),\sum_{k=0}^{4m-2}{2k\choose k}m^k\equiv4m-1\pmod{(4m-1)^2},

then 4m14m-1 is prime. If

k=04m(2kk)(m)k4m+1(mod(4m+1)2),\sum_{k=0}^{4m}{2k\choose k}(-m)^k\equiv4m+1\pmod{(4m+1)^2},

then 4m+14m+1 is prime. The authors report checking m1500m\leq1500 without finding counterexamples, while presenting this as a sufficient condition for primality.

Sources & referencesView supporting material

Primary source

Victor J. W. Guo and Jiang Zeng, “Some congruences involving central q-binomial coefficients”, arXiv:0910.3563 (2010).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.