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: http://euler.free.fr/download.htm   Unzip, run it. Click on Options/Uninstall Then Ctrl-N, and enter: power=5 start=0 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

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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
+15271462245795840^5+15304804584652800^5.

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

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