Problems & Puzzles: Puzzles

Puzzle 158.  Sum of Cubes equal to Square of Sum

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:

13 + 23 + 33 + ... + k3 = (1 + 2 + 3 + ... + k)2

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

a3 + b3 + c3 + ... + z3 = (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:

23 + 23 + 43 + 43+ .. + (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?

• 3125639893 (prime!)
• 33+13+23+53+63+33+93+83+93+33 =2401= (3+1+2+5+6+3+9+8+9+3)2

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

a3 + 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 (20 digits)

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 counterexample.

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

Solution:

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.

***

Frank Rubin sent (October, 2004) the solution to the second question:

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

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