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$$1, 3, 12, 32,...$$

Above is the sequence of the number of solutions, if there are, to the Diophantine equation : $d^{m+1} =a^{m}+ b^{m}+ c^{m}$ for $m =2$, in positive integers where $a, b$ and $c$ are realtively prime.

For example with the notation: $d^{3} =a^{2}+b^{2}+c^{2} --> (a,b,c,d)$

$$ [1, 1, 5, 3]\\ [3, 19, 31, 11] [9, 17, 31, 11] [19, 21, 23, 11]\\ [1, 19, 139, 27] [1, 71, 121, 27] [5, 83, 113, 27] [7, 95, 103, 27] [11, 49, 131, 27] [17, 25, 137, 27] [23, 55, 127, 27] [25, 37, 133, 27] [29, 41, 131, 27] [43, 47, 125, 27] [43, 85, 103, 27] [59, 89, 91, 27]\\ [1, 5, 207, 35] [1, 75, 193, 35] [1, 93, 185, 35] [1, 135, 157, 35] [3, 29, 205, 35] [5, 57, 199, 35] [9, 137, 155, 35] [11, 27, 205, 35] [15, 103, 179, 35] [15, 143, 149, 35] [19, 67, 195, 35] [25, 43, 201, 35] [25, 111, 173, 35] [27, 89, 185, 35] [33, 95, 181, 35] [33, 115, 169, 35] [37, 59, 195, 35] [43, 135, 151, 35] [47, 129, 155, 35] [51, 55, 193, 35] [51, 125, 157, 35] [55, 103, 171, 35] [61, 123, 155, 35] [65, 97, 171, 35] [65, 137, 141, 35] [67, 75, 181, 35] [71, 103, 165, 35] [73, 135, 139, 35] [79, 97, 165, 35] [79, 115, 153, 35] [103, 125, 129, 35] [109, 113, 135, 35] $$

So, we have 32 different forms of writing $42875$ as the sum of three squares; but only 1 form of expressing $27$ this way, 3 ways to express $11^3$ and 11 to write $3^9$. The sequence seems clearly to be infinite.

And here is the sequence for $m = 3$. It probably still is an infinte sequence :

$$1, 1, 1, 1, 1,..$$

But the sequences for $m \ge 4$ might well be nonexistent. For $m=4$ there might be, perhaps, a seldom very few ones.

The question is, as there always must be a question; find a "1" to $m =4$ and to $m =5$; that is.

P.S : Computing was made using the efficient Pari gp.

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What about the sequence $3, 11, 27, 35, \ldots$? –  Hagen von Eitzen Jan 25 '13 at 14:12
    
Yes $3,11,27,35,... $is another sequence we can built on the same question and for m =2. I was more interested in the number of solutions than in the solutions. –  user55514 Jan 25 '13 at 14:20

1 Answer 1

The sequence of the number of solutions seems to be : $1, 3, 12, 32, 32, 65, 64, 113, 62, 134,... $ for respective values of d: $3, 11, 27, 35, 51, 59, 75, 83, 99, 107,... $

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Neither of these is in OEIS yet, but compare oeis.org/A186301 = 3, 11, 27, 35, 51, 59, 99, 107, ... –  Hagen von Eitzen Jan 26 '13 at 12:25
    
Yes, the d´s are all 3 mod 8, but i do not know yet why. A 3 mod 8 cubed is 3 mod 8. But squares are 0, 1 or 4 mod 8 and the sum of three of them could be 0+1+4 = 5 mod 8 whose cubic root is 5 mod 8. Also cubes can be 0 mod 8, but not 2, 4 or 6 mod 8 and even numbers squared are 0 or 4 mod 8 and so i do not know why we do not get even d´s for sum of 3 squares: 0+0+0 = 0 mod 8 or 4+4+0 = 0 mod 8 and 4+0+4 = 0 mod 8 and 0+4+4 = 0 mod 8. –  user55514 Jan 26 '13 at 15:57
    
Well , i answer my own comment. I had forgotten the condition that a,b and c were relatively prime. Since 0 mod 8 and 4 mod 8 are even numbers, two of them (nor three of them) cannot be together to give a 5 mod 8 d^3 and also d result or a 0 mod 8 d^3 and also d result. Also we shall never have a 4+4+1 = 1 mod 8 d^3 or d result. Only 3 mod 8 (1+1+1 mod 8) d^3 and d results are possible. –  user55514 Jan 28 '13 at 17:50
    
And no way to get a d =7 mod 8 even though we released the realtively prime condition among a,b, and c. –  user55514 Jan 28 '13 at 18:21

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