Diophantine equation 23

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$${Published}$$  $${Online}$$  $${First}$$  $${(12/2/2024)}$$
$${Latest}$$  $${additions}$$  $${(12/2/2024)}$$
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$${Diophantine}$$ $${equation}$$ $${23}$$
$${(23.1)}$$  $${~~~\sum\limits_{i=1}^4\ a_i^5}$$ $${=}$$ $${\sum\limits_{j=1}^4\ b_j^5}$$
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$${case(23.1)}$$  $${~\sum\limits_{i=1}^4\ a_i^5}$$ $${=}$$ $${\sum\limits_{j=1}^4\ b_j^5}$$

まず次の様において代入すると、
$${a_1=ax+p~~,~~a_2=bx+q\\a_3=cx+r~~,~~a_4=dx+s\\b_1=ex+t~~,~~b_2=fx+u\\b_3=gx+v~~,~~b_4=hx+w}$$

$${x^5(a^5+b^5+c^5+d^5−e^5−f^5−g^5−h^5)\\+5x^4(a^4p+b^4q+c^4r+d^4s-e^4t-f^4u-g^4v-h^4w)\\+10x^3(a^3p^2+b^3q^2+c^3r^2+d^3s^2-e^3t^2-f^3u^2-g^3v^2-h^3w^2)\\+10x^2(a^2p^3 + b^2q^3 + c^2r^3 + d^2s^3 − e^2t^3 − f^2u^3 − g^2v^3 − h^2w^3)\\+5x(ap^4+bq^4+cr^4+ds^4−et^4−fu^4−gv^4−hw^4)\\+(p^5+q^5+r^5+s^5−t^5−u^5−v^5−w^5) = 0}$$
となる。
ここで、
$${(a^5+b^5+c^5+d^5−e^5−f^5−g^5−h^5)=0}$$
$${(p^5+q^5+r^5+s^5−t^5−u^5−v^5−w^5)=0}$$
となる様に$${(x^5~,~x^0}$$の項が$${0)}$$

$${(e,f,g,h)→(a,b,c,d)}$$
$${(t,u,v,w)→(q,r,s,p)}$$に置き換える。
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$${a_1=ax+p~~,~~a_2=bx+q\\a_3=cx+r~~,~~a_4=dx+s\\b_1=ax+q~~,~~b_2=bx+r\\b_3=cx+s~~,~~b_4=dx+p}$$
これを代入すると、$${x^5~,~x^0}$$の項がなくなり、

$${5x^4\{p(a^4−d^4)+q(b^4−a^4)+r(c^4−b^4)+s(d^4−c^4)\}\\+10x^3\{p^2(a^3−d^3)+q^2(b^3−a^3)+r^2(c^3−b^3)+s^2(d^3-c^3)\}\\+10x^2\{p^3(a^2-d^2)+q^3(b^2-a^2)+r^3(c^2-b^2)+s^3(d^2-c^2)\}\\+5x\{p^4(a-d)+q^4(b-a)+r^4(c-b)+s^4(d-c)\} = 0}$$

ここで$${x\ne0}$$とし、両辺を$${5x}$$で割ると
$${x^3\{p(a^4−d^4)+q(b^4−a^4)+r(c^4−b^4)+s(d^4−c^4)\}\\+2x^2\{p^2(a^3−d^3)+q^2(b^3−a^3)+r^2(c^3−b^3)+s^2(d^3-c^3)\}\\+2x\{p^3(a^2-d^2)+q^3(b^2-a^2)+r^3(c^2-b^2)+s^3(d^2-c^2)\}\\+\{p^4(a-d)+q^4(b-a)+r^4(c-b)+s^4(d-c)\} = 0}$$
となる。更に$${x^1~,~x^0}$$が$${0}$$になる様に、つまり

$${\{p^4(a-d)+q^4(b-a)+r^4(c-b)+s^4(d-c)\} = 0}$$
$${\{p^3(a^2-d^2)+q^3(b^2-a^2)+r^3(c^2-b^2)+s^3(d^2-c^2)\}=0}$$

$${1}$$つ目を$${a}$$について解くと
$${a=-\frac{b(q^4-r^4)+c(r^4-s^4)+d(s^4-p^4)}{p^4-q^4}}$$
$${2}$$つ目を$${a^2}$$について解くと
$${a^2=-\frac{b^2(q^3-r^3)+c^2(r^3-s^3)+d^2(s^3-p^3)}{p^3-q^3}}$$

$${\Bigr\{-\dfrac{b(q^4-r^4)+c(r^4-s^4)+d(s^4-p^4)}{p^4-q^4}\Bigr\}^2}$$

$${=-\dfrac{b^2(q^3-r^3)+c^2(r^3-s^3)+d^2(s^3-p^3)}{p^3-q^3}}$$

ここで$${b^2}$$の係数を比べると、
$${\bigg(\dfrac{q^4-r^4}{p^4-q^4}\Bigg)^2=-\bigg(\dfrac{q^3-r^3}{p^3-q^3}\Bigg)}$$
$${q=0~,~r=p}$$の時に成り立つ。
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$${q=0~,~r=p}$$を代入し$${b}$$で解くと
$${\Biggr\{-\dfrac{b(-p^4)+c(p^4-s^4)+d(s^4-p^4)}{p^4}\Biggr\}^2}$$

$${=-\dfrac{b^2(-p^3)+c^2(p^3-s^3)+d^2(s^3-p^3)}{p^3}}$$

$${→\Biggr\{b-c\Bigg(1-\dfrac{s^4}{p^4}\Bigg)-d\Bigg(\dfrac{s^4}{p^4}-1\Bigg)\Biggr\}^2}$$

$${=b^2-c^2\Bigg(1-\dfrac{s^3}{p^3}\Bigg)-d^2\Bigg(\dfrac{s^3}{p^3}-1\Bigg)}$$

$${s/p=t}$$とおくと
$${\bigr\{b-c\big(1-t^4\big)-d\big(t^4-1\big)\bigr\}^2}$$
$${=b^2-c^2\big(1-t^3\big)-d^2\big(t^3-1\big)}$$

$${→\bigr\{b-(c-d)(1-t^4)\bigr\}^2}$$
$${=b^2-(c^2-d^2)(1-t^3)}$$

$${→b^2-2b(c-d)(1-t^4)+(c-d)^2(1-t^4)^2=b^2-(c^2-d^2)(1-t^3)}$$

$${b=\dfrac{(c-d)^2(1-t^4)^2+(c^2-d^2)(1-t^3)}{2(c-d)(1-t^4)}}$$

$${b=\dfrac{(c-d)(1-t^4)}{2}+\dfrac{(c+d)(1+t+t^2)}{2(1+t)(1+t^2)}}$$

$${a=b-(c-d)(1-t^4)}$$に代入する。

$${a=-\dfrac{(c-d)(1-t^4)}{2}+\dfrac{(c+d)(1+t+t^2)}{2(1+t)(1+t^2)}}$$

$${a,b}$$がこの時$${x^1~,~x^0}$$が$${0}$$になる。
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$${x^3\{p(a^4−d^4)+q(b^4−a^4)+r(c^4−b^4)+s(d^4−c^4)\}\\+2x^2\{p^2(a^3−d^3)+q^2(b^3−a^3)+r^2(c^3−b^3)+s^2(d^3-c^3)\}=0}$$
$${x^2}$$で割り$${q=0~,~r=p}$$を代入する
$${x\{p(a^4−d^4)+p(c^4−b^4)+s(d^4−c^4)\}\\+2\{p^2(a^3−d^3)+p^2(c^3−d^3)+s^2(d^3-c^3)\}=0}$$
$${s/p=t}$$とすると
$${x=-2p\Bigg(\dfrac{(a^3−b^3)+(c^3−d^3)(1-t^2)}{(a^4−b^4)+(c^4−d^4)(1-t)}\Bigg)}$$
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ここまでを整理すると、
$${a_1=ax+p~~,~~a_2=bx+q\\a_3=cx+r~~,~~a_4=dx+s\\b_1=ax+q~~,~~b_2=bx+r\\b_3=cx+s~~,~~b_4=dx+p}$$が解であり、

$${q=0~,~r=p~,~s/p=t}$$

$${a=-\dfrac{(c-d)(1-t^4)}{2}+\dfrac{(c+d)(1+t+t^2)}{2(1+t)(1+t^2)}}$$

$${b=\dfrac{(c-d)(1-t^4)}{2}+\dfrac{(c+d)(1+t+t^2)}{2(1+t)(1+t^2)}}$$

$${x=-2p\Bigg(\dfrac{(a^3−b^3)+(c^3−d^3)(1-t^2)}{(a^4−b^4)+(c^4−d^4)(1-t)}\Bigg)}$$

$${(c,d,p,s)}$$を決めると上記から求まる。
$${(c,d,p,s)=(1,2,1,-2)}$$の時

$${(a,b,t,x)=\bigg(-\dfrac{42}{5}, \dfrac{33}{5},-2, \dfrac{7160}{12651}\bigg)}$$
これら全てを代入し分母を払うと、

$${a_1=47256~~,~~a_2=19811\\a_3=60144~~,~~a_4=18142\\b_1=47493~~,~~b_2=59907\\b_3=10982~~,~~b_4=26971}$$
と解が求まる。
$${47256^5+19811^5+60144^5+18142^5\\=47493^5+59907^5+10982^5+26971^5}$$
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$${REFERENCES}$$
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$${\small【Narinder~Kumar~Wadhawan】}$$
$${\footnotesize SOLUTION~~TO~EQUAL~~SUM~~OF\\FIFTH~~POWER~~DIOPHANTINE\\EQUATIONS~–~A~~NEW~~APPROACH}$$
$${\small J\tilde{n}\bar{a}n\bar{a}bha,~Vol.~53(1) (2023), ~~125-145}$$
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【目次001】・【目次002】・【目次003】
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