# Quintic diophantine equation

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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One (trivial) solution is $a=b=c=d=3.$ –  Jonathan Christensen Jan 24 at 2:02
Also $a=b=c=d=0$. –  Mario Carneiro Jan 24 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 at 2:08
@ABlumenthal: You mean not equal. –  André Nicolas Jan 24 at 2:31
@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 at 6:23

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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You are heartless. –  Will Jagy Jan 24 at 2:23
On the other hand, this may be all solutions. –  Will Jagy Jan 24 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 at 12:03
hahaha @GerryMyerson, +1 –  Rustyn Jan 27 at 2:01

Relatively prime may be difficult:

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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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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 at 3:49

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

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So, no examples with coprime terms. –  Gerry Myerson Jan 24 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 at 3:35
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 at 4:23
Which powers are we summing over? Can you please elaborate more? –  Ethan Jan 24 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 at 7:13
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 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 $e^{3}+ f^{3}+ 3^2g^{3} =3g^{4}$
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 at 12:18