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Background: The equation

$$a^4+b^4+c^4=2d^4$$

has infinitely many positive integral solutions if we take $c=a+b$ and $a^2+ab+b^2=d^2$ further assuming that $GCD(a,b,c)=1$.

Main problem: Find some positive integral solutions to the equation

$$a^4+b^4+c^4=2d^4$$

with $a\lt b\lt c\ne a+b$ and $GCD(a,b,c)=1$.

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    $\begingroup$ $a = 4014369822293252845298556668977$ $b = 8028813922964684804294250901215$ $c = 8331871536210073175631303374584$ $d = 8243128117136361914922521992201$ $\endgroup$
    – Tomita
    Commented Jan 12, 2023 at 3:40
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    $\begingroup$ Jovan, you need to accept the answer if you are satisfied with it. $\endgroup$
    – user25406
    Commented Jan 12, 2023 at 13:39
  • $\begingroup$ @user25406 I'm satisfied with the Tomita's answer. $\endgroup$
    – user178594
    Commented Apr 3, 2023 at 17:37
  • $\begingroup$ There are no solutions for $c\le300$. $\endgroup$
    – user178594
    Commented Apr 4, 2023 at 9:14
  • $\begingroup$ Has it been noted before that in the case of $c=a+b$ and $d^2=a^2+ab+b^2$ not only $a^4+b^4+c^4=d^4$ holds but also $a^2+b^2+c^2=2d^2$? However this property doesn't seem to hold for the solutions with $c\not=a+b$ given in this thread. $\endgroup$
    – Frederick Magata
    Commented May 2 at 5:58

2 Answers 2

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Let $A=\frac{a}{d}, B=\frac{b}{d}, C=\frac{c}{d},$ then we get

$$A^4+B^4+C^4=2\tag{1}$$

Let $A=x+y, B=x-y, z=C^2$ then

$$2x^4+12y^2x^2+2y^4+z^2 = 2\tag{2}$$

Hence

$$y^2 = -3x^2 \pm \frac{\sqrt{32x^4+4-2z^2}}{2}\tag{3}$$

So we find the rational solutions of $(4)$.

$$v^2=32x^4+4-2z^2\tag{4}$$

$(4)$ can be parameterized to $(5)$ using $(z,v)=(4x^2,2)$ with $w=(k^2+2)C$.

$$w^2 = 4(k^2+2)(-2+k^2)x^2-4(k^2+2)k\tag{5}$$

Thus we must find the rational solutions of simultaneous equations $(3),(5)$.

First we get a parametric solution of $(5)$ using giving some rational number $k$.

Take $k=\frac{-9}{13}$. Then we get:

$$x = \frac{3}{142}\frac{28561m^2-430732-7081100m}{28561m^2+430732}$$ $$w = \frac{-1257}{11999}\frac{-10768300+173732m+714025m^2}{28561m^2+430732}$$ $$C = \frac{-3}{71}\frac{-10768300+173732m+714025m^2}{28561m^2+430732}$$

Hence $(3)$ becomes

$$y^2 = \frac{10083247442281m^4-41255619857608m^2+2788930240200m^3+2293337020040464-42060204482400m}{20164(28561m^2+430732)^2}$$

Hence we must find the rational solutions $(m,V)$ of $(6)$
$$V^2 = 10083247442281m^4+2788930240200m^3-41255619857608m^2-42060204482400m+2293337020040464\tag{6}$$

Quartic equation $(6)$ is birationally equivalent to the elliptic curve using a known solution $(m,V)=(360/169,-47445892)$ as follows. $$Y^2 = X^3 -X^2 -1097465452X+ 3288951361780\tag{7}$$

Though I couldn't find the generator of elliptic curve , I got one solution $P(X,Y)=(\cfrac{-97636631990}{5536609}, \cfrac{53963430434179560}{13027640977})$.

Since $(7)$ has rank $1$, we get infinitely many solutions of $(7)$ using group law as follows.

For instance, we get a solution of $(6)$,
$2Q(m,V)=(\cfrac{-762617488871059540}{36474714629699307},\cfrac{-63980049963485293932709632365019549028}{46581170383319934672021751209})$

We get $(x,y)=(\cfrac{4317501713916820330332746705607}{16486256234272723829845043984402},\cfrac{12346241358503326020929860043561}{16486256234272723829845043984402})$

Finally, we get

$a = 4014369822293252845298556668977$
$b = 8028813922964684804294250901215$
$c = 8331871536210073175631303374584$
$d = 8243128117136361914922521992201$

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  • $\begingroup$ Elegant, but why do you suppose this solution is so far from minimal? See the other answer. $\endgroup$
    – Oscar Lanzi
    Commented Jan 15, 2023 at 10:55
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    $\begingroup$ I tried to make it understandable without much prerequisite knowledge. $\endgroup$
    – Tomita
    Commented Jan 15, 2023 at 13:11
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    $\begingroup$ If everyone were a professional mathematician, maybe $10$ lines answer might be enough. $\endgroup$
    – Tomita
    Commented Jan 15, 2023 at 23:57
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    $\begingroup$ @OscarLanzi: Maybe has to do with the fact it was not brute-force but a rational point on an elliptic curve. Quite similar to this MSE post about co-prime solutions to $a^4+b^4+c^4 = 9d^2$ where an elliptic curve yields $$2682440^4+15365639^4+18796760^4 = 9\times141668657747643^2$$ but the smallest by brute force is only $$155^4+260^4+296^4 = 9\times37747^2.$$ $\endgroup$
    – Tito Piezas III
    Commented Feb 6, 2023 at 10:12
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    $\begingroup$ If we take the point $Q(m)=\frac{360}{169}$ then we get a small solution $(a,b,c,d)=( 32, 1065, 2321, 1973).$ The large solution corresponds to the point $2Q(m)=\cfrac{-762617488871059540}{36474714629699307}.$ $\endgroup$
    – Tomita
    Commented Feb 7, 2023 at 4:09
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$(a, b, c, d) = (32, 1065, 2321, 1973), (2156, 5605, 8381, 7383)$

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    $\begingroup$ did you use an algorithm or brute force? $\endgroup$
    – user25406
    Commented Jan 10, 2023 at 13:31
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    $\begingroup$ @user25406 brute force $\endgroup$
    – Peter Mueller
    Commented Jan 10, 2023 at 13:38
  • $\begingroup$ More primitive ones: (9845, 44747, 78212, 67467); (20091, 58120, 115003, 98267); (54796, 76165, 172667, 146907); (37028, 64555, 209731, 176799); (72681, 156145, 207512, 187589); (122213, 246996, 303115, 280531); (157131, 167560, 324691, 281239); (116745, 155873, 575528, 484813). The values of $d$ in primitive irregular solutions are 1973, 7383, 67467, ... (OEIS A121995), defined as ``Denominators of rational points on $x^4+y^4+z^4=2$ not satisfying $z=x+y$''. Found by simple search algorithm. $\endgroup$
    – Rosie F
    Commented Jun 3 at 13:47

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