Problems & Puzzles: Puzzles Puzzle 103. N=a4+b4
= c4+d4 635318657
is the least number that is the sum of two biquadrates in two different
ways. 635318657=
594+1584 = 1334 + 1344 But
635318657 is not prime. Question 1. Find the least and
two more examples where N,
a & c
are prime
(if they exist... now I doubt about their existence) If
we want that all the four a, b, c & d numbers are primes (of course
that in this case N is even), this is the
least example: 3262811042
= 74+2394
= 1574+2274 Question
2. Find 3 more examples of these. Question 3. Any shortcut to find them (1 & 2)? Question 4. A fresh interpretation of the title of the background song by Robin Frost? Refs. pp. 275 & 290-291, "Recreations in The Theory of Numbers", A. H. Beiler, (except for the 4th question)Solution Question 1 Chris Nash wrote "N cannot be prime. Because a prime of the form 4x+1 can only be expressed as a sum of two squares in exactly one way. (and fourth powers are of course examples of squares)" *** Question 2 Jean-Charles Meyrignac points out (14/08/2000) that "D.J. Bernstein already explored equations a^4+b^4 = c^4+d^4. You can find the 516 first solutions at http://cr.yp.to/sortedsums/two4.1000000 (detected broken 1/9/01) *** In his page Bernstein says also that "218 of the solutions were found previously by Lander, Parkin, Selfridge, and Zajta". With the help of a little code in Ubasic I (C.R.) analyzed the 516 solutions as DATA statements and found only other prime solution: 620474 + 403514 = 596934 + 467474 = 17472238301875630082 *** Stuart Gascoigne wrote (2/6/04): The following numbers are all prime.... 157662524204984267584824067340911^4 + 175518862296574361970383530947991^4 = 174460113691778517117959086988941^4 + 159090599632575616653359441298211^4 269758686329508376469851929750719^4 +
534065979590887060090455886663777^4 = 1679539956802461427023806692932554869^4+2864939822128245005298014063613916133^4= 2735636962517662175684582548040182073^4+2096549864621014042130013452441703601^4 I took this identity f2(a,b)=-a^13+a^12*b+a^11*b^2+5*a^10*b^3+6*a^9*b^4-12*a^8*b^5-4*a^7*b^6+7*a^6*b^7-3*a^5*b^8-3*a^4*b^9+4*a^3*b^10+2*a^2*b^11-a*b^12+b^13 f2(a,b)^4+f2(b,-a)^4=f2(a,-b)^4+f2(b,a)^4 which was originally published as "Some new results on equal sums of like powers", Simcha Brudno Mathematics of computation volume 25 1969 pp. 877-880. I found it on Jean-Charles Meyrignac's site http://euler.free.fr/identities.htm.I then tried all values of a<b<1000 and tested the values produced to see if they were prime. I did this as a spreadsheet using Joe Crump's ZZMath Excel addin http://www.spacefire.com/numbertheory/default.asp?SubPage=ZMath.htm.This gave me some values which are 'probable primes'. I then used Primo by Marcel Martin http://www.ellipsa.net/ to check they were prime.The entire computation took maybe one hour and gave me 4 results. In the tradition of these things, I conjecture that there are an infinite number of prime solutions. ***
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