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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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    $\begingroup$ One (trivial) solution is $a=b=c=d=3.$ $\endgroup$
    – Jonathan Christensen
    Jan 24, 2013 at 2:02
  • $\begingroup$ Also $a=b=c=d=0$. $\endgroup$
    – Mario Carneiro
    Jan 24, 2013 at 2:05
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    $\begingroup$ If you apply the sum-of-three-squares theorem, then you can show that if $(a,b,c,d)$ is any solution, then $d = 4^k(8m+7)$ where $k, m \geq 0$. This is not sufficient, however, but narrows the search somewhat. $\endgroup$
    – A Blumenthal
    Jan 24, 2013 at 2:12
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    $\begingroup$ @ABlumenthal: You mean not equal. $\endgroup$
    – André Nicolas
    Jan 24, 2013 at 2:31
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    $\begingroup$ @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) $\endgroup$
    – Eric Naslund
    Jan 24, 2013 at 6:23

6 Answers 6

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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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    $\begingroup$ You are heartless. $\endgroup$
    – Will Jagy
    Jan 24, 2013 at 2:23
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    $\begingroup$ On the other hand, this may be all solutions. $\endgroup$
    – Will Jagy
    Jan 24, 2013 at 2:29
  • $\begingroup$ 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). $\endgroup$
    – Adam Bailey
    Jan 24, 2013 at 12:03
  • $\begingroup$ hahaha @GerryMyerson, +1 $\endgroup$
    – Rustyn
    Jan 27, 2013 at 2:01
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We can use the identity,

$$(2p-2q)^4+(2p+2q)^4+(4q)^4 = 2^5(p^2+3q^2)^2\tag1$$

One can then solve,

$$p^2+3q^2 = (a^2+3b^2)^k$$

for any $k$. For $k=5$, it is,

$$p =a^5 - 30 a^3 b^2 + 45 a b^4$$ $$q=b (5 a^4 - 30 a^2 b^2 + 9 b^4)$$

though $(1)$ has the common factor $2$.

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  • $\begingroup$ @Ethan: Hey, you deleted that nice comment. I was about to reply to it. Anyway, it's wonderful to know that others like my "Collection of Algebraic Identities" and I appreciate hearing about it. I enjoyed creating it and I hope others enjoy it too. Thanks. :) $\endgroup$
    – Tito Piezas III
    Jan 12, 2018 at 7:59
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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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    $\begingroup$ 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 $\endgroup$
    – coffeemath
    Jan 24, 2013 at 3:49
  • $\begingroup$ Solutions with gcd=2 exist: $a,b,c,d=124,174,298,98$ $\endgroup$
    – elluser
    Jun 9, 2014 at 13:01
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$k=1000;for(a=1,k,for(b=a,k,for(c=b,k,if(ispower(a^4+b^4+c^4,5,&n),print([a,b,c,n]))))) [3, 3, 3, 3] [14, 252, 266, 98] [18, 18, 36, 18] [22, 44, 77, 33] [33, 66, 66, 33] [83, 83, 249, 83] [96, 96, 96, 48] [98, 196, 294, 98] [110, 110, 176, 66] [124, 174, 298, 98] [163, 489, 489, 163] [226, 226, 339, 113] [356, 534, 534, 178] [729, 729, 729, 243]$

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    $\begingroup$ So, no examples with coprime terms. $\endgroup$
    – Gerry Myerson
    Jan 24, 2013 at 5:39
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a parametric equation that generates solutions?

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

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  • $\begingroup$ 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. $\endgroup$
    – Ethan Splaver
    Jan 24, 2013 at 3:35
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    $\begingroup$ 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. $\endgroup$
    – zyx
    Jan 24, 2013 at 4:23
  • $\begingroup$ Which powers are we summing over? Can you please elaborate more? $\endgroup$
    – Ethan Splaver
    Jan 24, 2013 at 6:20
  • $\begingroup$ 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. $\endgroup$
    – zyx
    Jan 24, 2013 at 7:13
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    $\begingroup$ 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? $\endgroup$
    – Gerry Myerson
    Jan 24, 2013 at 12:19
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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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    $\begingroup$ The important difference between $m=3$ and $m=4$ is $(1/3)+(1/3)+(1/3)+(1/4)\gt1$ whereas $(1/4)+(1/4)+(1/4)+(1/5)\lt1$. $\endgroup$
    – Gerry Myerson
    Jan 25, 2013 at 12:18
  • $\begingroup$ 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 ? $\endgroup$
    – user55514
    Jan 26, 2013 at 13:45
  • $\begingroup$ And yes: 11^4 +29^4 + 26^4 + 37^4 = 19^5 with Sum >1. $\endgroup$
    – user55514
    Jan 26, 2013 at 14:18
  • $\begingroup$ 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. $\endgroup$
    – user55514
    Jan 26, 2013 at 14:33
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    $\begingroup$ 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. $\endgroup$
    – Gerry Myerson
    Jan 26, 2013 at 23:47

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