Problems & Puzzles: Puzzles

Puzzle 697 Puzzle 105, revisited

From our almost 13 years old Puzzle 105, we know that a prime fifth power as a sum of k prime powers, the minimal k found was 7, by Jean-Charles Meyrignac on September 13, 2000:
2843=2731+1933+1459+709+509+271+31 (Jean-Charles Meyrignac, 09/13/2000)

Two more solutions are know of this type, one smaller and another larger, discovered some time after:

1709=1567+1373+719+503+431+367+349 (Michael Lau, 08/20/2002)
4027=3301+3169+3037+2411+1481+859+569 (Takao Nakamura, 2/10/2008)

Meyrignac himself wrote in the year 2000:

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

Perhaps is time to see if we can get now a prime solution using five primes in the right part:

f^5 =a^5+b^5+c^5+d^5+e^5, where a, b, c, d, e & f are prime numbers... (E1)

Q1. Find a prime solution to E1.

The best approximation that I have produced to the asked question Q1 is this one:

72^5 = 67^5 + 47^5 + 46^5 + 43^5 + 19^5 (4/6 primes)

Q2. Perhaps you can get a better approximation if you do not succeed with Q1.

Note: If you are interested in the free J-C code to save time,

Just download the program here:
Unzip, run it. Click on Options/Uninstall
Then Ctrl-N, and enter:
end=the largest term you want
Left terms=1
Right terms=5
Search: Exact
Click on OK, and the computation will start.
All the solutions are saved into the file SOLUTION.TXT.


Contributions came from Seiji Tomita

Q2: I used Sastry's identity as follows.
(u^5+75*v^5)^5=(u^5+25*v^5)^5+(u^5-25*v^5)^5+(-u^5+75*v^5)^5+(10*u^3*v^2)^5 +(50*u*v^4)^5

Substitute {u,v}={1501, 672} to above identity, we obtain,
17897072968929901^5= 11045088504648301^5+4193104040366701^5+2658880423914899^5

4/6 primes: {17897072968929901,11045088504648301,4193104040366701,2658880423914899}



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