To Frank Rubin
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:
+ 23 + 33 + ... + k3 = (1 + 2 + 3 +
... + k)2
this time I started thinking in the associated Diophantine equation:
+ b3 + c3 + ... + z3 = (a + b + c + ... + z)2
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.
trial and error I discovered that there is another family of solutions
using only even numbers, the following way:
+ 23 + 43 + 43+ .. + (2k)3
+(2k)3 = (2 + 2 + 4 + 4 + ... + 2k + 2k)2
question: Do you devise any other family of solutions, easy to write
down as these two above?
the before digressions still do not configure a prime-puzzle.
started thinking about prime numbers (zeros free!) such that their
digits - taken one by one - satisfy the Diophantine equation stated
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.
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
The answer is negative. Example:
- 3125639893 (prime!)
At this point I was seeking for solutions to the Diophantine equation
+ b3 + c3 + ... + z3 = (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
Second question: prove that there are finite
solutions to a3
+ b3 + c3 + ... + z3 = (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
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:
Set k=2 and you get the serie you have found. For k=3 (i.E.) you can
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).
The trivial solution is n=0 (this means, you always can continue with
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
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
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