Problems & Puzzles: Puzzles

Puzzle 102. Three squares a cube, all primes

This means: A^2+B^2+C^2 = D^3, A<=B<=C, {A, B, C & D} = primes

Here are a few examples I have gotten: 

3²+3²+3²=3³
3²+19²+31²=11³
3²+691²+2293²=179³
3²+5869²+54959²=1451³
3²+24967²+60169²=3²+28163²+58741²=1619³
3²+13127²+121229²=2459³

Questions:
1. Find other examples.
2.  Can A be other than 3?

Solution

Jud McCranie found (4-5/8/2000) the explanation for the question 2 and other solutions to question 1, including another cube with double expression. Here are them:

"If p is prime then p=2, p=3, or p^2 = 1 mod 6. If A^2 = 1mod 6 then A^2+B^2+C^2 = 3 mod 6, which doesn't work since the cube of a prime > 3 is = 1 or = 5 mod 6. If A=2, 2^2+B^2+C^2, B>A is ruled out with a parity argument. A=B=2 is ruled out since 2^2+2^2+C^2 = 3 mod 6. Therefore A must be 3".

***

Chris Nash has given (7/8/2000) a qualitative jump in this search discovering a way of seeking candidates to cubes expressible as sum of 3 squares: D must be 3 modulo 8; then he shows a methodic way of looking for D^3-9 as a sum of squares. As a consequence of his method he could find plenty of new solutions and a very interesting example of D: "In particular, 90659^3 is the sum of three prime squares in FIVE different ways!.

Here is his email:

"Jud McCranie found the first solutions to this problem, now I’d like to add more information - and an algorithm to find some much larger solutions!

The first part was already proven by Jud. One of the three numbers on the left-hand side must be 3. Hence we seek numbers D such that D^3-9 is a sum of two squares. Since the square of all odd numbers is 1 modulo 8, D must be of the form 8x+3. So now we need to do the following test

• loop for (prime) values of D, starting at 3, and with step 8
• try to express D^3-9 as a sum of two (prime) squares.

To perform this (last) calculation, we will quote the following results.

Any prime of the form 4x+1 can be expressible as the sum of two squares (a result known to Fermat). Primes of the form 4x+3 are certainly not a sum of two squares.

The prime 2 can be expressed as 1^2+1^2.

If N=a^2+b^2, then k^2N = (ka)^2+(kb)^2

If M=a^2+b^2 and N=c^2+d^2, then MN=(ac+bd)^2+(ad-bc)^2. (This rule can be applied several ways, by changing the signs of a, b, c, d or changing the order of the terms).

It can be shown these rules are complete. In other words, using these rules we can find all ways of expressing D^3-9 as a sum of two squares:

• Factor D^3-9 (it will be twice an odd number). If there are any factors of the form 4x+3, exit, unless D^3-9 is twice the square of a prime.  (That gives only the result 3 3 3 3).
• Express each prime factor as a sum of two squares.  Use the product rule in all possible combinations to generate all possible sums of two squares that sum to D^3-9.
• Check for primes.

Using this method I tested for all values up to D=2^(53/3), about 200000, a limit imposed by double-precision arithmetic. It was possible to factor or primality-test all values of D^3-9, using PrimeForm/GW for any difficult cases. Each prime factor was expressed as a sum of two squares by an exhaustive search, not the most efficient algorithm, but good enough.

Here are the complete results for A, B, C, D. These confirm Jud’s results and go much further - several more prime cubes were found that were the sum of three prime squares in more than one way. In particular, 90659^3 is the sum of three prime squares in FIVE different ways!

Best wishes

Chris

Results (listed in ascending order of D)

...
3 752449 1654019 14891
3 479581 1913861 15731
3 101021 2271439 17291
3 963863 2492851 19259
3 1987429 2287063 20939
3 1076827 2860021 21059
3 485753 3155687 21683
3 2415913 3455561 26099
3 529127 4241759 26339
3 2924567 4207837 29723
3 943421 5477867 31379
3 1244183 5417479 31379
3 2134963 5270549 31859
3 1943861 5412437 32099
3 2704019 5280371 32771
3 3707021 5062403 34019
3 4771463 7229969 42179
3 1161871 8944679 43331
3 3544427 8563727 44123
3 236129 9394129 44531
3 2054707 9861707 46643
3 1585253 10583773 48563
3 1949881 10716857 49139
3 1983257 11159789 50459
3 7443613 8974409 51419
3 7654007 9226109 52379
3 6874201 10472647 53939
3 6718469 11394289 55931
3 8934637 9846217 56123
3 5715323 13238663 59243
3 7863293 12087007 59243
3 11942101 15315071 72251
3 8951737 17692901 73259
3 12365567 15637487 73523
3 9931007 19416611 78059
3 7307303 22541041 82499
3 837379 24356777 84059
3 9676741 23085611 85571
3 3597527 26145313 88643
3 1535441 26907191 89891
3 8647391 25526009 89891
3 4284971 26958677 90659
3 4891079 26855327 90659
3 5216759 26793967 90659
3 6989461 26387093 90659
3 12200333 24418909 90659
3 17998501 21739967 92699
3 5329699 27715817 92699
3 17202139 25322357 97859
3 18302021 24862009 98411
3 15802517 28650449 102299
3 16058831 28507573 102299
3 7708627 33720853 106163
3 10804769 34498111 109331
3 15535747 33262169 110459
3 23057479 30097427 112859
3 5889467 39117899 116099
3 27484153 30226579 118619
3 17979173 37320377 119723
3 24001723 34511531 120899
3 24903077 34720039 122219
3 23583877 37362301 124979
3 1995661 46580333 129539
3 28362031 37574813 130379
3 29568863 36966499 130859
3 10370531 46361171 131171
3 27429881 40890257 134339
3 30764729 39539693 135899
3 31247507 40277023 137483
3 411119 53491729 141971
3 20647513 49612657 142403
3 8672423 56507833 148403
3 37255061 45672239 151451
3 4206703 60667183 154643
3 8662783 60335767 154883
3 7645817 61389169 156419
3 2344787 62175559 157019
3 23507089 58997347 159179
3 17479879 62854027 162059
3 19759771 63317677 163859
3 26829163 66527627 172643
3 17790637 70033553 173483
3 43306699 57879791 173531
3 21868501 70825913 176459
3 43803757 61668689 178859
3 10436291 75272971 179411
3 21435941 73654927 180539
3 35732023 77166799 193379
3 36684601 76718527 193379
3 39108983 75957229 193979
3 49505893 69142309 193379
3 29215301 81974177 196379
3 44971517 77317027 200003"

***

Using the Chris Nash method I search for one solution being D a palprime, 3 mod 8 of course, and such that D^3-9 = 2*Prime, in order to simplify the calculation of the sum of the squares. Not far I found this one:

3²+5857398397²+1168183957² = 3291923³

But I do not pretend that this is the least palprime that satisfies the equation. Would you look for it?

BTW maybe you'll find interesting the following code to run under Ubasic, by Malm, for expressing a prime P = 4k+1 as a sum of two squares.

10 input P
20 if P@4<>1 then A=0:B=0:print "P wrong":end
30 Sqrtp=isqrt(P):if (res=0) then A=0:B=0:return endif
40 T=1:repeat inc T until kro(T,P)=-1
50 G=modpow(T,(P-1)\4,P):if P\2<G then G=P-G endif
60 A=P@G
70 if A=1 then B=G else
80 :while Sqrtp<A:Te=G:G=A:A=Te@G wend
90 :if A=0 then B=0 else:B=G@A endif endif
100 print A,B

***

Mahmoud Chilali wrote (9/8/2000):

***

Regarding the question tied by Chilali, Said el aidi wrote (18/8/2000):

"The equation posed by Mahmoud Chilali is only a specific case of Mordell's equation y² = x³ + k (k<> 0), Baker proved in 1968: If x, y are integers and y² = x³ + k, then

                 max{|x| , |y|} < exp { (10¹⁰ * |k|)^10000 } 

which means that the equation y2 = x3 + k has at most finitely many solutions in integers. The exponent still must be reduced before we can use computers to take care of all the small cases.

For more details, the interested reader will be anble to consult the book of Baker A. “The diophantine equation y2 = x3 + k “ Phil. Trans. R. Soc. London, 263, 1968, 193-208"

***