To **Frank Rubin**

The
last weekend I was digging around killing my stress reading at random my
Mathematical Handbook of Formulas and Tables, **Schaum's** series, and one
more time I fell in captivity with the marvelous equation:

**1**^{3}
+ 2^{3} + 3^{3} + ... + k^{3} = (1 + 2 + 3 +
... + k)^{2}

But
this time I started thinking in the associated Diophantine equation:

**a**^{3}
+ b^{3 } + c^{3} + ... + z^{3} = (a + b + c + ... + z)2

The
first question that I made myself was if the series of the natural numbers from 1 to k
(a=1, b=2, ... z=k) was the only family of solutions.

By
trial and error I discovered that there is another family of solutions
using only even numbers, the following way:

**2**^{3}
+ 2^{3} + 4^{3} + 4^{3}+ .. + (2k)^{3}
+(2k)^{3} = (2 + 2 + 4 + 4 + ... + 2k + 2k)^{2}

**First
question: Do you devise any other family of solutions, easy to write
down as these two above?**

But
the before digressions still do not configure a prime-puzzle.

Then I
started thinking about **prime numbers (zeros free!) such that their
digits - taken one by one - satisfy the Diophantine equation stated
above.**

The
least prime of this type is **1423**. As
you can recognize, this prime has digits that correspond to the first
family of solutions: the natural numbers from a=1 to k=4.

Evidently
the second family of solutions (composed of only even digits, twice
each) can not compose a prime number. So the following question was: **are
all the primes of this type just permutations of the first k<=9
natural numbers?**

The answer is negative. Example:

- 3125639893 (prime!)
- 3
^{3}+1^{3}+2^{3}+5^{3}+6^{3}+3^{3}+9^{3}+8^{3}+9^{3}+3^{3}
=2401= (3+1+2+5+6+3+9+8+9+3)^{2}

At this point I was seeking for solutions to the Diophantine equation

**a**^{3}
+ b^{3} + c^{3} + ... + z^{3} = (a + b + c + ...
+ z)^{2 }............( I )

such that all the values a, b, c, ... & z are positive numbers
less than 10 and that - additionally - these values concatenated compose
a prime number.

The next question I tackled was to ask if I could get primes of these
type of __ any length__, which is analog than to ask if these kind of primes are
finite or infinite.

To my surprise I discovered (how?) that there is a
limit to the possible length for these primes; moreover, __ I
conjecture__ that the largest prime of this kind is **99151111111111111111
**(20 digits)

**Second question: prove that there are finite
solutions to ****a**^{3}
+ b^{3} + c^{3} + ... + z^{3} = (a + b + c + ...
+ z)^{2}** using only positive numbers less than 10 **

**Third question: Prove that 99151111111111111111
is the largest prime of the type discussed here, or get a
counterexample.**

**Fourth question: Find examples of these
primes with K digits for K=15, 16, 17, 18 & 19**** **

**Rudolph Knjezk** has found (28/10/01) a general solution to the
asked in first question:

There is an infinity of such series. There is a constructive proof:

1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2

k^4*(1^3+2^3+3^3+...+n^3=k^4*(1+2+3+...+n)^2

k*(k^3+(2k)^3+(3k)^3+...(nk)^3)=(k*(k+2k+3k+...nk))^2

Set k=2 and you get the serie you have found. For k=3 (i.E.) you can
write:

3^3+3^3+3^3+6^3+6^3+6^3+9^3+9^3+9^3+...(3n)^3+(3n)^3+(3n)^3 =
(3+3+3+6+6+6+9+9+9+...3n+3n+3n)^2

He also sent the following observation:

Furthermore I have found a remarkable property of the original serie.
Let k be the sum of numbers and k^2 the sum of their third powers. We want
to continue the serie by one new element (n).

k^2+n^3=(k+n)^2

k^2+n^3=k^2+2kn+n^2

n^3=2kn+n^2

The trivial solution is n=0 (this means, you always can continue with
0)

n^2-n-2k=0

n=(1+sqrt(1+8k))/2

For a positive integer n (1+8k) must be the square of an odd number. If
you start with k=0 the first n=1 and the new k=1. k=1 allows an expansion
only with n=2 and the new k=3. We get only the original series. This
means, there is no other seried fitting the Diophantine equation, than can
be expanded ad infinitum with one by one element. Maybe, this is a little
helpful to anyone, who tries to find an answer to your other questions.

Suppose there are N terms in each sum. Then the maximum
value of the left side is (9^3+9^3+...+9^3)=729N. The minimum value of
the right side is (1+1+...+1)^2=N^2. Therefore, for N>729 the right
side will always be greater than the left side. So no solutions with
N>729 exist.

He has also calculated by his own the largest number
following the equation (I): 99511111111111111111, result that responds
indirectly the third question. 99151111111111111111 is the largest
prime of this type.

BTW the number found by **Rubin** is a composite
number divided by 7.

**Rubin** explains how he found his number:

The technique is fairly simple. I have 9 nested
loops. The loop variables represent the number of 9's, 8's, 7's,
etc., in the number. I calculate the sum of the cubes and the square
of the sum for each partial number. If the square is larger than the
sum of the cubes, I terminate that loop.

Once I get down to the number of 2's, I approximate the
number of ones that will be needed, and loop from that point up. It
usually takes just 2 iterations of the innermost loop, that is, the
1's loop.

At the end, I just count the total number of digits.
If that is the best so far, then I print it out. It is not necessary
to form and compare the numbers because of the way the loops are
nested

***