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.
On
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
=======================
On
a parametric equation that generates solutions?
The expected number of integer solutions without common factor is finite, so no.
On
$k=1000;for(a=1,k,for(b=a,k,for(c=b,k,if(ispower(a^4+b^4+c^4,5,&n),print([a,b,c,n]))))) [3, 3, 3, 3] [14, 252, 266, 98] [18, 18, 36, 18] [22, 44, 77, 33] [33, 66, 66, 33] [83, 83, 249, 83] [96, 96, 96, 48] [98, 196, 294, 98] [110, 110, 176, 66] [124, 174, 298, 98] [163, 489, 489, 163] [226, 226, 339, 113] [356, 534, 534, 178] [729, 729, 729, 243]$
On
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}$
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....