Problems & Puzzles: Puzzles

Puzzle 105. a^5 = b1^5+b2^5+...+bn^5

Jean-Charles Meyrignac proposes the following puzzle:

We know that 144^5=133^5+110^5+84^5+27^5 (found by Lander, Parkin and Selfridge in 1966). In how much terms a prime at the power 5 can be expressed as a sum of other primes at the power 5 ? In other words, what is the minimum n in the following equation: a^5 = b1^5+b2^5+...+bn^5 where a,b1,b2,...,bn are all prime numbers ?

As probably you know Jean Charles has an interesting site in the Web coordinating the search for records of those Diophantine equations called "Sum of Like Powers", ( ).

As a matter of fact, in his pages he provides to the interested people in these records, a code - sharply named "Euler2000" - of his own to make the search, and ranges to cover with it.

Jean Charles has promised modify his current code for readers who desire to use it for the purposes of this puzzle, not without "...the hope someone else will write a better program...". According to an email received this morning (Saturday, August 26) the modified code will be available since tomorrow Sunday.

Question 1: Find the asked nmin value and the corresponding particular example.

Before posting this proposed puzzle I wanted to know if any solution using only primes is ever possible. Handling this issue is that I (C.R.) can state the following true statement:

"Exist at least one solution using only primes, and therefore nmin<2166902"

Question 2: Can you tell us where came from my statement?

Another interesting questions that I would like to append is the following one:

Question 3: Find any solution using a set of consecutive primes for the b quantities.

Question 4: Find any solution using a set of consecutive and all distinct primes for the b quantities.


Carlos Rivera found (26/08/2000) one solution to question 3 just digging around with a very elemental code in Ubasic (while the Euler2000 for primes arrive):

7^5 = 5^5+38*3^5+139*2^5

As you can see this solution involves not only consecutive primes for the "b" quantities but for all the bases employed  ("a" & "b's"). As a sub-product of this little search the nmin has been slow downed to 178.

But of course that I do not pretend that this is the shorter solution using consecutive primes... would you try other shorter ones?


Regarding the original question 1, and after posting the prime-version of the Euler2000, Jean-Charles has gotten the following starting solution to be improved by the readers:

(5,1,9):  431^5=421^5+229^5*2+173^5+151^5*2+149^5+67^5+61^5

As you can see this solutions has only 7 distinct prime in the right side of the equation. But with this preliminary solution, certainly nmin has been slow downed to 9

Jean-Charles says "...for the moment, (5,1,4) has only one solution, so I think we can reach (5,1,5) with prime values..."


A solution of nine terms with all distinct primes in the right side, is the following found by Jean-Charles him-self:


Who adds, "...But I'm still sure that someone can find a better solution with less terms !..."


Chris Nash has produced (30/8/2000) some interesting results and has outlined a method to do this search. Inside the method a questions arises for the readers...

"Hola Carlos, I thought I would attempt Puzzle 105. Here is my reasoning.

37 = 5 mod 2^5.
37^5 = 5^5 mod 2^5

Hence 37^5 - 5^5 is divisible by 2^5 and so 37^5 = 5^5 + 2166901*2^5, which is where you got your original bound, n<=2166902.

[Well, as a matter of fact I simply postulated that probably P1^5-P2^5= k*2^5 for some P1 & P2<P1; then I made an exhaustive search with the known result, C.R.]

Now we note that 2166901 = 17^5 + 2*13^ 5 + 5^5 + 5*3^5 + 118 And so by rearranging we have 37^5 = 5^5 + 32 * (17^5+2*13^5+5^5+5*3^5) + 118*2^5 =
32 * 17^5 + 64*13^5 + 33*5^5 + 160*3^5 + 118*2^5 (*), which gives n=407.

A little later I discovered the interesting result: 7^5 = 61*3^5 + 62*2^5. (n=123).

[A nice result!.. consecutive primes and consecutive factors, at the right side of the equation...C.R.]

Here is an elementary search I performed on Puzzle 105. As a result, I reduce n_min to 77, and the example I show also uses consecutive primes (in fact, the same consecutive primes).

First the result.

7^5 = 3*5^5 + 24*3^5 + 50*2^5, n=77

And now the method.

The Diophantine equation Ax+By=C where A, B and C are known integers, can be solved in integers if G=gcd(A,B) divides C. The solution method is easy, once we realize that we can solve Ax+By=G by using Euclid’s gcd algorithm. We note also that the solution is not unique - in fact there are an infinity of solutions. (Given any solution, add B/G to x, and subtract A/G from Y). By multiplication we can get a solution for C. Now to use this algorithm. We generate a list of primes up to 2^(10.6), this limit is chosen so double-precision floating point will not overflow when computing fifth powers. For each pair of primes p>q>3, calculate the positive numbers p^5-k.q^5 for integer values of k. Assign this value to C, let A$3, and B2. Solve the Diophantine equation, and of the infinity of solutions, seek one where x>=0, y>=0, and x+y is minimal. Then p^5 is partitioned into k+x+y prime fifth powers. A quick search discovers the solution very quickly.

Of course, n_min must be *much* smaller than 77 - can anyone else do better by using more distinct primes, or not using the primes 2 and 3?  I believe that by collecting more and more fifth power sums, and adding and subtracting, better results can be obtained. For instance, by replacing multiples of 3*5^5+24*3^5+50*2^5 in (*), the value of n in that equation can be reduced. (I think I can see a method, but it seems *very* inefficient!)."

[Maybe it's still inefficient but a method is a magic stick against exhaustive search...isn't it?, C.R ]


The 31/8/2000 Chris Nash improved his own solution for the consecutive case:

13^5=4.2^5+17.3^5+9.5^5+7^5+2.11^5, n=33

He is using a program from his own and expects to beat the (5,1,9) solution obtained by Jean-Charles.


Jean Charles improves (4/9/2000) the solution for consecutive primes from n=33 to n=22:


37^5 = 31^5 + 29^5 + 23^5*2 + 19^5*2 + 17^5 + 13^5*2 + 11^5 + 7^5*3 + 5^5 + 3^5*7 + 2^5


Other people has been using the Jean-Charles code to search solutions with primes close related to the matter of this puzzle:

(5,2,6) 719+491=739+191+113+101+47+19 (Michael Lau)

(5,3,5) 103+83+17=109+41+31+19+3 (Torbjörn Alm, 09/05/2000)

(5,4,4) 101+97+43+29=109+73+71+17 (Torbjörn Alm, 09/05/2000)

Torbjörn Alm also checked that no solution to (5,1,8) exists below 2750^5. Jean Charles says "I believe that (5,1,7) is reachable, but when ?"


The 'when' was the Wednesday 13 of September, 2000. Jean Charles wrote that day:

I just found: (5,1,7) 2843=2731+1933+1459+709+509+271+31 !!!

I don't think that (5,1,5) exists, but who knows ? Perhaps there is a proof of its non-existence ?




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