Puzzle 702 Adding emirps to a cube

Contributions came from J. K. Andersen, Farideh Firoozbakht, Torbjörn Alm and Abhiram R Devesh

Andersen wrote:

The 11 smallest pairs have sum 705628^3:
160409920439929091 + 190929934029904061
160519903159918091 + 190819951309915061
160519933129918091 + 190819921339915061
160619902259917091 + 190719952209916061
160719953109916091 + 190619901359917061
160809944019925091 + 190529910449908061
160819900459915091 + 190519954009918061
160919930429914091 + 190419924039919061
170819920439915081 + 180519934029918071
170909931329924081 + 180429923139909071
170909934029924081 + 180429920439909071

***

Farideh wrote:

It seems that " if both numbers n & R(n) are odd then n+R(n) can not be
a cube ". So there is no any emip p such that p+R(p) is a cube and there is
no any solutions for this puzzle... I haven't any proof now. But I feel confident that it is true. I will think on it....
I also think, 4 is the only positive integer n such that n & R(n) are even
and n+R(n) is a cube. In fact it seems that if n+R(n) is even then there is
no nontrivial solution for n+R(n) = m^3. Namely it seems that (x, y) = (1, 4)
is the only solution of the equation 8x^3 = y+R(y).
So I think your puzzle 702 hasn't any solution... of course. The proof isn't easy for me.

***

Torbjörn wrote:

I have scanned up to 10^12.

No prime solution found. A prime solution implies that prime+emirp is even and
divisable by 8.

All solutions are grouped with a common cube solution.
Solutions with 9 digit primes: 35 solutions with cube root=707 followed by >100 solutions with cube root = 1019
Solutions with 10 digit primes: 30 solutions with cube root =2321
Solutions with 11 digit primes: >1000 solutions with cube root = 4891

***

Abhiram wrote:

I could not find any solution p less than 23870331827

***

Later, on August 30, 2013, Andersen wrote:

My search identified cubes which might be the sum of a number p and its reversal R(p).
If there are no carries in the addition then p+R(p) is a palindrome, since both the first and last digit of p+R(p) is the sum of the first and last digit of p, and so on. A carry can only increase a digit by 1 or turn 9 into 0, so we can test whether a cube is "close" enough to being a palindrome. There are additional conditions to get carries in the right places and get potential for emirps but the details are omitted here.
When promising cubes were found, another program tried all ways to write them as p+R(p).

As mentioned, the first 11 pairs have sum 705628^3
The next 35 pairs have sum 34343434^3, starting with:
10000090001030201070503 + 30507010203010009000001

There are no other solutions below 10^33.

***