Problems & Puzzles: Puzzles

Puzzle 436. 43 =4^2+3^3

JM Bergot poses the following question:

Q1. Are there many primes like 43=4^2 + 3^3?

You are able to use any exponent (greater than 0) for each digit to get the sum of these powers to equal the original prime.

Carlos Rivera found many other examples as the asked by Bergot. In order to reduce the cases I added the following conditions;

a) The number composed by the exponents must be prime too
b) No zero-digit is allowed neither in the prime base (P) nor in the prime exponent (Q)

43=4^2 + 3^3, 23 is prime too, (P, Q ) = (43, 23)

Other example: (3947, 7351): 3947 = 3^7+9^3+4^5+7^1

Q2. Get other interesting curio prime numbers like these, for large P primes.

 

Giovanni Resta wrote:

Hi Carlos,
here are my initial findings for Puzzle 436, maybe I'll add
something during the week.

The number of primes which satisfy the conditions of Q1 or Q2
is finite, since 10*9^9 < 10^10, so these primes can have
at most 10 digits.

The 3 largest primes which satisfy Q1 and Q2 are the same.
The couples are:
(1594994699, 2999996689)
(1693979849, 3991949939)
(1693979869, 8994949919)
Note that there can be several exponents for
the same base.
The maximal prime 1693979869 can also have prime exponent 8194949999.
The largest couple with a palindromic prime base is
(394696493, 469178399) and there are no pairs in which
the exponent is palindromic too.

...

I have little to add to my previous e-mail about Puzzle 436.
I have computed all the zeroless prime pairs which satisfy condition
Q2. They are 155970. I had no time to examine them, but if
anybody is interested, the pairs can be retrieved from:
http://ilex.iit.cnr.it/~resta/p436.txt  (about 3,000 Kb)
or http://ilex.iit.cnr.it/~resta/p436.gz  (about 670 Kb

***

Jan van Delden wrote:

The number of solutions to this puzzle is finite.

The maximum number of digits (prime or no prime) of P,Q is 10 (and maybe lower). Assuming we have a number P and Q of the form {9}n=10^n-1, the sum is equal to n*(9^9). This must (at least) equal P itself. Solving n*(9^9)=(10^n-1) gives n<10, which gives n=10 by rounding above. So 11 digits or more is impossible.

In order to get at 10 digits, well need a fair amount of 8s and 9s in P and 9s in Q, but our resulting sum (P) will start with a small digit, so these 8s and 9s are concentrated at the end of P. Furthermore P should be prime, so possibly the maximum could be set a bit lower. Analyzing these kind of equations in greater detail could restrict the set of combined digits (P-Q) for every n (or at least several n).

***

 

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