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How can I find non trivial primitive integer solutions, to the Diophantine equation $$a^4+b^4+c^4=d^5$$ Can anyone find me solutions to this equation?

Or if possible a parametric equation that generates solutions?

I would appreciate any help

Ive also simplified it to finding coprime integer solutions greater then 1 to the equation,$$xyz(x^2+y^2+z^2)=1250w^5$$ I don't know if that helps at all.

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2  
One (trivial) solution is $a=b=c=d=3.$ –  Jonathan Christensen Jan 24 '13 at 2:02
    
Also $a=b=c=d=0$. –  Mario Carneiro Jan 24 '13 at 2:05
    
@MarioCarneiro That doesn't satisfy $a,b,c,d > 1$, although the notation for that requirement is a bit ambiguous. –  Erick Wong Jan 24 '13 at 2:08
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@ABlumenthal: You mean not equal. –  André Nicolas Jan 24 '13 at 2:31
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@GerryMyerson: I don't think so, because our equation here is not homogeneous. I just thought it may be relevant, and that perhaps there was a way to use fourth powers to get a fifth power. (Although that sounds far fetched) –  Eric Naslund Jan 24 '13 at 6:23

5 Answers 5

Pick any three numbers, say $1,2,3$. Compute $1^4+2^4+3^4=1+16+81=98$. Multiply through by $98^4$, and voila! $$98^4+196^4+294^4=98^5$$ If you insist on relatively prime solutions, you may have to work a little harder....

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4  
You are heartless. –  Will Jagy Jan 24 '13 at 2:23
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On the other hand, this may be all solutions. –  Will Jagy Jan 24 '13 at 2:29
    
Very neat solution which generalizes to some other Diophantine equations in mixed powers, eg for $a^4 + b^4 + c^4 = d^7$, multiply through by 98 to the power of 20 (20 being cong 0 mod 4 and cong -1 mod 7). –  Adam Bailey Jan 24 '13 at 12:03
    
hahaha @GerryMyerson, +1 –  Rustyn Jan 27 '13 at 2:01

Relatively prime may be difficult:

=======================

d       a       b       c
0       0       0       0
1       0       0       1
2       0       2       2
3       3       3       3
16       0       0      32
17       0      17      34
18      18      18      36
32       0      64      64
33      22      44      77
33      33      66      66
48      96      96      96
66     110     110     176

=======================

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3  
The case of $(22,44,77;33)$ isn't exactly obtained by Myerson's remark, however $2^4+4^4+7^4=2673=11\cdot 3^5$, so multiplying each by 11 gives the required other factor of $11^4$ to bring the prime power of 11 up to a multiple of 5. There may be something in this idea of generating all solutions by perturbing results of summing three random fourth powers. +1 –  coffeemath Jan 24 '13 at 3:49
    
Solutions with gcd=2 exist: $a,b,c,d=124,174,298,98$ –  elluser Jun 9 at 13:01

$k=1000;for(a=1,k,for(b=a,k,for(c=b,k,if(ispower(a^4+b^4+c^4,5,&n),print([a,b,c,n]))))) [3, 3, 3, 3] [14, 252, 266, 98] [18, 18, 36, 18] [22, 44, 77, 33] [33, 66, 66, 33] [83, 83, 249, 83] [96, 96, 96, 48] [98, 196, 294, 98] [110, 110, 176, 66] [124, 174, 298, 98] [163, 489, 489, 163] [226, 226, 339, 113] [356, 534, 534, 178] [729, 729, 729, 243]$

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2  
So, no examples with coprime terms. –  Gerry Myerson Jan 24 '13 at 5:39

a parametric equation that generates solutions?

The expected number of integer solutions without common factor is finite, so no.

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What do you mean "expected number of integer solutions", there are plenty of parametric equations in multiple variables capable of generating co prime solutions, to similar Diophantine equations. –  Ethan Jan 24 '13 at 3:35
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There is a probabilistic argument that if the sum of 1/(degree of term using each variable) is less than $1$, the number of coprime solutions is finite. When the sum is larger than $1$ the approximate number of solutions less than $n$ predicted by the same argument is consistent with a polynomial parametrization. –  zyx Jan 24 '13 at 4:23
    
Which powers are we summing over? Can you please elaborate more? –  Ethan Jan 24 '13 at 6:20
    
The degrees of $a^4, b^4, c^4$ and $d^5$ are 4,4,4, and 5. The sum in question is (1/4 + 1/4 + 1/4 + 1/5) which is less than 1. –  zyx Jan 24 '13 at 7:13
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The number of sums of three 4th powers up to $N$ is, roughly, $N^{3/4}$, so the probability that a random number less than $N$ is a sum of three 4th powers is roughly $N^{-1/4}$. The number of 5th powers uo to $N$ is $N^{1/5}$, so the probability that some 5th power is a sum of three 4th powers is roughly $N^{1/5}N^{-1/4}=N^{-1/20}$ which goes to zero as $N\to\infty$. This is the probabilistic argument. It's not a proof, just a heuristic, but it seems to be quite reliable. Why are people voting zyx down? Is it just out of ignorance? –  Gerry Myerson Jan 24 '13 at 12:19

Have a look at this experimental result with Pari gp for $a^{m}+b^{m}+c^{m} = d^{m+1}$ and m =3. $? k=1000;for(a=1,k,for(b=a,k,for(c=b,k,if(ispower(a^3+b^3+c^3,4,&n)&gcd(a,b)==1&gcd(a,c)==1&gcd(b,c)==1,print actor(n)]))))) [19, Mat([19, 1]), 89, Mat([89, 1]), 117, [3, 2; 13, 1], 39, [3, 1; 13, 1]] [75, [3, 1; 5, 2], 164, [2, 2; 41, 1], 293, Mat([293, 1]), 74, [2, 1; 37, 1]] [81, Mat([3, 4]), 167, Mat([167, 1]), 266, [2, 1; 7, 1; 19, 1], 70, [2, 1; 5, 1; 7, 1]] [107, Mat([107, 1]), 163, Mat([163, 1]), 171, [3, 2; 19, 1], 57, [3, 1; 19, 1]] [222, [2, 1; 3, 1; 37, 1], 263, Mat([263, 1]), 961, Mat([31, 2]), 174, [2, 1; 3, 1; 29, 1]] [225, [3, 2; 5, 2], 362, [2, 1; 181, 1], 407, [11, 1; 37, 1], 106, [2, 1; 53, 1]] [323, [17, 1; 19, 1], 333, [3, 2; 37, 1], 433, Mat([433, 1]), 111, [3, 1; 37, 1]] [397, Mat([397, 1]), 441, [3, 2; 7, 2], 683, Mat([683, 1]), 147, [3, 1; 7, 2]]$

We can see there are primitive solutions if m <4 because there are probably some hidden identities; i am myself skilless and have not enough mathematical knowledge to find them; but one seems to be $ e^{3}+ f^{3}+ 3^2g^{3} =3g^{4}$

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The important difference between $m=3$ and $m=4$ is $(1/3)+(1/3)+(1/3)+(1/4)\gt1$ whereas $(1/4)+(1/4)+(1/4)+(1/5)\lt1$. –  Gerry Myerson Jan 25 '13 at 12:18
    
Well, Sum(1/exponents) more or less than one, has relatively prime solutions or not; i understand it. But isn´t it another arbitrary barrier/bordier experimentally found; (as if it was physics and not mathematics)? In other words why is it ? How can we explain/prove it ? –  user55514 Jan 26 '13 at 13:45
    
And yes: 11^4 +29^4 + 26^4 + 37^4 = 19^5 with Sum >1. –  user55514 Jan 26 '13 at 14:18
    
And yes: 11^4 +29^4 + 26^4 + 37^4 = 19^5 with Sum >1. And also a beautiful Guy "law of small numbers" result, because: For the sum of a 4, almost all of 4 odd, fourth powers equal to a fifth odd power , one of them must be even. And with this minimal solution we do have p^4 + q^4 + (2r)^4 +s^4 = t^5 with all p,q,r,s,t prime. –  user55514 Jan 26 '13 at 14:33
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There is nothing arbitrary or experimental about the distinction based on the sum of the reciprocals of the exponents. It is based on probabilistic reasoning, as indicated in the comments below the answer from @zyx. –  Gerry Myerson Jan 26 '13 at 23:47

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