Problems & Puzzles: Puzzles

Puzzle 228. Sum of twin primes, a square

Let p & q to be twin primes (q=p+2) such that the sum of p & q is a square, that is to say, such that p + q =x2

Example p = 17, q = 19, p + q = 36 = 62

Question: Find the four earliest titanic p & q twins having such property*.

_____
* Hint: The last four digits of the earliest titanic p are 7799. 


Solution:

Faride Firoozbakht and J. K. Andersen solved this puzzle.

Faride sent the smaller prime of the earliest prime:

p1= 100000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000004035547918766213409
143725517651500630631615470558308725599978102680556354957
647888154089895525475555728384686970883323297896566371618
048846796405280294448103718577575188283753156232217118620
241083258464603199204322161923387677962200724322044068968
814279448736951000247607913059170052171767482013869497090
860662129497784722507443360087889195705901906846852947317
589502799647198831609727809825063794946341897892861163188
850925128857335975617279386988149030265106919106546335830
9465685383859118461101308257799

She also sent one more solutions.

a=Ceiling[((10^999+1)/2)^(1/2)] p=2(a+n)^2-1 , q=2(a+n)^2+1. If n >= 0 and p & q are primes, then (p,q) is such pair of twin primes. p1=2(a+451187)^-1 (I wrote before). p2=2(a+1004390)^2-1

***

J. K. Andersen wrote:

First 5 prp solutions:

18*(x+floor(sqrt(5)*10^499/3))^2+/-1, for x=150396, 334797, 639583, 905665, 950299. The prp's have not been proved. The smallest ends with 7799 as the hint says. The solutions were found with my own sieve to 2^31, and PrimeForm/GW for prp testing.

He added:

The complexity of twin prp's with N digits is near O(N^4). The earliest 2000-digit prp solution: 18*(550423+floor(sqrt(5)*10^999/3))^2+/-1

***

 

 

 



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