Problems & Puzzles: Puzzles

Let’s define the following kind of "Palprimes" (palindromic primes) : those that are the sum of the same power of consecutive numbers

Ppal = x ⁿ +(x+1) ⁿ

1. Can you find the first 5 of them for each even power ?
2. ( a minor question) why there are not any primes for odd powers ?

Solution

Jud. McCranie sent us a mail (Fri. 17 Jul. 1998)

"For n=2: 

x	Ppal
1	5
9	181
12	313
1262	3187813

 and the 5th one has eluded me. for n=4: x=9, Ppal = 16561

For the part b) of this puzzle he wrote :

"Pal-primes and sum of powers… - why not any primes for odd powers? 

Because if n is odd then x^n + (x+1)^n is divisible by 2x+1. If n > 1 then 2x+1 is a proper divisor of x^n+(x+1)^n, so it isn't prime. There are, of course, solutions when n=1".

***

Patrick De Geest also wrote us (Sat, 18 Jul 1998) for the same subject :

"I visited puzzle page 014 (Palprimes and sum of powers). Did you know I started a search palindromic sums of squares of two consecutive numbers ? The largest one I found ( unfortunately not a prime)is the following 17 digit number : 80.472.264^2 + 80.472.265^2 = 12951570707515921. The next three ones (181,313,3187813) are palindromic primes :

9^2 + 10^2 = 181
12^2 + 13^2 = 313
1262^2 + 1263^2 = 3187813

So, only two more to find, but it will be difficult ! I didn't check for other powers. Two years of delving into palindromic numbers revealed to me that palindromes in general don't thrive in higher dimensions, so I think that solutions in higher than 2 powers will be very, very hard to find".