Problems & Puzzles: Puzzles

Puzzle 804. Palindromes and primes

Somewhere out there I saw this question: "find prime P such that (P+1)^2+1=Q is another prime and Q^2+Rev(Q^2) is palindrome R

I made a preliminary search and found that P=19 is the smallest solution where R=268862

But more than that I found a total of 38 solutions very easily being the last P=1616689 where R=7853587698556558967853587, before I stopped my code.

Thinking in these big quantity of solutions I offer to you the following questions.

Q1. 38 solutions in this range of primes (19 to 1616689) are really too many or not.

Q2. Why all of these 38 prime solutions end in digit 9?

Q3. May R be a prime number? Show it.


Contributions came from Dmitry Kamenetsky, Jan van Delden and Emmanuel Vantieghem


Dmitry wrote:

I found 508 solutions for P<=5,000,000,000. In this range there were 5 solutions with prime R:
P=49513019, R=7090069339891712171989339600907
P=157999999, R=723201296000829989928000692102327
P=3874728899, R=325429994494888298949892888494499924523
P=3911806909, R=336759449495657632747236756594944957633
P=4998501539, R=726554258873768478545874867378852455627
All P end with 9, but I am not sure why.


Jan wrote:

Q3: Solutions with p,q,r prime:
49513019 2451539149520401 7090069339891712171989339600907
157999999 24964000000000001 723201296000829989928000692102327
3874728899 15013524048495210001 325429994494888298949892888494499924523
3911806909 15302233301123748101 336759449495657632747236756594944957633
4998501539 24985017645382371601 726554258873768478545874867378852455627
13129010299 172370911457506090001 39719852246692648777677784629664225891793
50442302609 2544425892598812812101 7498405479957477558580858557747599745048947
A property of these solutions is that there is no carry-over in the computation of r=q^2+Rev(q^2).
Note my routine did not use this result (otherwise it could be much faster), does this always happen?


Emmanuel wrote:

I found  826 primes < 2*10^10  that satisfied the conditions of the problem.
They all end with a 9.
It is clear that  p  cannot end with 1 or 7 because then  q  ends in  5.
But there can be very big primes ending with 3 that satisfied the conditions of the problem.
That these cases are very sparse is due to the fact that the corresponding value q end in 7, thus  q^2  ends in 9; numbers m  ending in 9 for which  m + R(m)  is palindrome are very rarely squares.  In fact, actually I don't know of such an  m.

For the following primes p, the  r  is prime :
49513019, 157999999, 3874728899, 3911806909, 4998501539, 13129010299, 17306619409... (their number is probably infinite).




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