Problems & Puzzles: Puzzles

 Puzzle 768. Primes as m sums of triangular numbers. Zak Seidov sent the following nice puzzle:   5657745031, 6676776031, 7313189731, 9717526081, 9835772131,11007657781,11375689081, 12602729701, 13365484681, 14115702331, 14762690881, 14822294581.  Search stopped at 16347754531.No results for m=100. These  prime numbers (all == 1 mod 30), have  m=96 representations as sum of two triangular numbers. Q1. Find more such numbers with as large m's as possible. Q2. Are there such "record" prime numbers not == 1 mod 30? Q3. Why this 1 mod 30 is exceptional.? Q4. Find smallest primes with m = {1 .. 1000} (very hard?) Q5 Why large numbers of m are (almost all) even? Find primes with large odd m's

Contributions came from Seiji Tomita,

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Seiji wrote:

Q1:
P=x(x+1)/2+y(y+1)/2
Let X=2x+1, Y=2y+1, then X^2+Y^2=8P+2.

Sums of two squares formula(Legendre?)
Let N=2^t*p1^(e1)*p2^(e2)*p3^(e3)*...*pn^(en)
{p1,p2,..,pn} mod 4=1
Representing number: R(N)=4(e1+1)*(e2+1)*(e3+1)*...*(en+1)

By using above formula, we get the unique representing number m= R(N)/8.
We consider the number N=(2)*(5)^e1*(13)^e2*(17)^e3*(29)^e4*(37)^e5*(41)^e6*(53)^e7.
e1,e2,..,e7<10

For example, maximum number in this range is m=3240000.
m=3240000
N=(2)*(5)^9*(13)^9*(17)^9*(29)^8*(37)^9*(41)^7*(53)^8
P=484054168542421269680458548814220935128074345469284924875418191598980535644531

Q2,Q3:
Primes such that P=1 mod 30 are not exceptional.
For example, P=13 mod 30, m=100.
m=100
N=(2)*(13)*(17)^4*(37)*(41)^4*(53)
P=1504153182893533

Q5:
Only one of {e1,e2,...,en} must be 1 and remaining all must be even.

m=81
N=(2)*(5)*(13)^2*(17)^2*(29)^2*(37)^2
P=70290074611

m=675
N=(2)*(5)^4*(13)^2*(17)^4*(29)^2*(37)^2*(53)
P=134579134102564531

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