Problems & Puzzles: Puzzles
Puzzle 105. a^5 = b1^5+b2^5+...+bn^5
Jean-Charles Meyrignac proposes the following puzzle:
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", (http://euler.free.fr ).
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:
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...
I thought I would attempt Puzzle 105. Here is my reasoning.
= 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.
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
nice result!.. consecutive primes and consecutive factors, at the right
side of the equation...C.R.]
[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:
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:
(5,3,5) 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: