The problem of four cubes

The problem of four cubes is to find all integer solutions of the Diophantine equation :

x^{3}+y^{3}+z^{3}=w^{3}.

{\ displaystyle x ^ {3} + y ^ {3} + z ^ {3} = w ^ {3}.}

{\ displaystyle x ^ {3} + y ^ {3} + z ^ {3} = w ^ {3}.}

It should be noted that while several complete solutions of this equation in rational numbers have been proposed, its complete solution in integers for 2018 is unknown ^[1] .

Examples of integer solutions

The smallest natural solutions:

3^{3}+4^{3}+5^{3}=6^{3}

{\ displaystyle 3 ^ {3} + 4 ^ {3} + 5 ^ {3} = 6 ^ {3}}

3^{3}+4^{3}+5^{3}=6^{3}

1^{3}+6^{3}+8^{3}=9^{3}

{\ displaystyle 1 ^ {3} + 6 ^ {3} + 8 ^ {3} = 9 ^ {3}}

1^{3}+6^{3}+8^{3}=9^{3}

3^{3}+10^{3}+18^{3}=19^{3}

{\ displaystyle 3 ^ {3} + 10 ^ {3} + 18 ^ {3} = 19 ^ {3}}

3^3+10^3+18^3=19^3

7^{3}+14^{3}+17^{3}=20^{3}

{\ displaystyle 7 ^ {3} + 14 ^ {3} + 17 ^ {3} = 20 ^ {3}}

7^{3}+14^{3}+17^{3}=20^{3}

4^{3}+17^{3}+22^{3}=25^{3}

{\ displaystyle 4 ^ {3} + 17 ^ {3} + 22 ^ {3} = 25 ^ {3}}

4^{3}+17^{3}+22^{3}=25^{3}

18^{3}+19^{3}+21^{3}=28^{3}

{\ displaystyle 18 ^ {3} + 19 ^ {3} + 21 ^ {3} = 28 ^ {3}}

18^{3}+19^{3}+21^{3}=28^{3}

11^{3}+15^{3}+27^{3}=29^{3}

{\ displaystyle 11 ^ {3} + 15 ^ {3} + 27 ^ {3} = 29 ^ {3}}

11^{3}+15^{3}+27^{3}=29^{3}

2^{3}+17^{3}+40^{3}=41^{3}

{\ displaystyle 2 ^ {3} + 17 ^ {3} + 40 ^ {3} = 41 ^ {3}}

2^{3}+17^{3}+40^{3}=41^{3}

6^{3}+32^{3}+33^{3}=41^{3}

{\ displaystyle 6 ^ {3} + 32 ^ {3} + 33 ^ {3} = 41 ^ {3}}

6^{3}+32^{3}+33^{3}=41^{3}

16^{3}+23^{3}+41^{3}=44^{3}

{\ displaystyle 16 ^ {3} + 23 ^ {3} + 41 ^ {3} = 44 ^ {3}}

16^{3}+23^{3}+41^{3}=44^{3}

If negative values are allowed, then the following identities hold:

-1^{3}+9^{3}+10^{3}=12^{3}

{\ displaystyle -1 ^ {3} + 9 ^ {3} + 10 ^ {3} = 12 ^ {3}}

-1^{3}+9^{3}+10^{3}=12^{3}

-2^{3}+9^{3}+15^{3}=16^{3}

{\ displaystyle -2 ^ {3} + 9 ^ {3} + 15 ^ {3} = 16 ^ {3}}

-2^{3}+9^{3}+15^{3}=16^{3}

-2^{3}+15^{3}+33^{3}=34^{3}

{\ displaystyle -2 ^ {3} + 15 ^ {3} + 33 ^ {3} = 34 ^ {3}}

-2^{3}+15^{3}+33^{3}=34^{3}

-2^{3}+41^{3}+86^{3}=89^{3}

{\ displaystyle -2 ^ {3} + 41 ^ {3} + 86 ^ {3} = 89 ^ {3}}

-2^{3}+41^{3}+86^{3}=89^{3}

-3^{3}+22^{3}+59^{3}=60^{3}

{\ displaystyle -3 ^ {3} + 22 ^ {3} + 59 ^ {3} = 60 ^ {3}}

-3^{3}+22^{3}+59^{3}=60^{3}

Complete rational parameterization

G. Hardy and Wright (1938) ^[2] ^[3]

$x=-a(b-3c)(b^{2}+3c^{2})+a^{4}$ ${\ displaystyle x = -a (b-3c) (b ^ {2} + 3c ^ {2}) + a ^ {4}}$ $x=-a(b-3c)(b^{2}+3c^{2})+a^{4}$
$y=\quad a(b+3c)(b^{2}+3c^{2})-a^{4}$ ${\ displaystyle y = \ quad a (b + 3c) (b ^ {2} + 3c ^ {2}) - a ^ {4}}$ $y=\quad a(b+3c)(b^{2}+3c^{2})-a^{4}$
$z=\quad a^{3}(b-3c)-(b^{2}+3c^{2})^{2}$ ${\ displaystyle z = \ quad a ^ {3} (b-3c) - (b ^ {2} + 3c ^ {2}) ^ {2}}$ $z=\quad a^{3}(b-3c)-(b^{2}+3c^{2})^{2}$
$w=\quad a^{3}(b+3c)-(b^{2}+3c^{2})^{2}$ ${\ displaystyle w = \ quad a ^ {3} (b + 3c) - (b ^ {2} + 3c ^ {2}) ^ {2}}$ $w=\quad a^{3}(b+3c)-(b^{2}+3c^{2})^{2}$

N. Elkies ^[1]: ${\begin{cases}x=d(-(s+r)t^{2}+(s^{2}+2r^{2})t-s^{3}+rs^{2}-2r^{2}s-r^{3}),\\y=d(t^{3}-(s+r)t^{2}+(s^{2}+2r^{2})t+rs^{2}-2r^{2}s+r^{3}),\\z=d(-t^{3}+(s+r)t^{2}-(s^{2}+2r^{2})t+2rs^{2}-r^{2}s+2r^{3}),\\w=d((s-2r)t^{2}+(r^{2}-s^{2})t+s^{3}-rs^{2}+2r^{2}s-2r^{3})\end{cases}}$ {\ displaystyle {\ begin {cases} x = d (- (s + r) t ^ {2} + (s ^ {2} + 2r ^ {2}) ts ^ {3} + rs ^ {2} - 2r ^ {2} sr ^ {3}), \\ y = d (t ^ {3} - (s + r) t ^ {2} + (s ^ {2} + 2r ^ {2}) t + rs ^ {2} -2r ^ {2} s + r ^ {3}), \\ z = d (-t ^ {3} + (s + r) t ^ {2} - (s ^ {2} + 2r ^ {2}) t + 2rs ^ {2} -r ^ {2} s + 2r ^ {3}), \\ w = d ((s-2r) t ^ {2} + (r ^ { 2} -s ^ {2}) t + s ^ {3} -rs ^ {2} + 2r ^ {2} s-2r ^ {3}) \ end {cases}}}

Other Solution Series

Leonard Euler , 1740

$x=1-(a-3b)(a^{2}+3b^{2})$ ${\ displaystyle x = 1- (a-3b) (a ^ {2} + 3b ^ {2})}$
$y=-1+(a+3b)(a^{2}+3b^{2})$ ${\ displaystyle y = -1 + (a + 3b) (a ^ {2} + 3b ^ {2})}$
$z=-a-3b+(a^{2}+3b^{2})^{2}$ ${\ displaystyle z = -a-3b + (a ^ {2} + 3b ^ {2}) ^ {2}}$
$w=-a+3b+(a^{2}+3b^{2})^{2}$ ${\ displaystyle w = -a + 3b + (a ^ {2} + 3b ^ {2}) ^ {2}}$

Linnik , 1940

$x=b(a^{6}-b^{6})$ ${\ displaystyle x = b (a ^ {6} -b ^ {6})}$
$y=a(a^{6}-b^{6})$ ${\ displaystyle y = a (a ^ {6} -b ^ {6})}$
$z=b(2a^{6}+3a^{3}b^{3}+b^{6})$ ${\ displaystyle z = b (2a ^ {6} + 3a ^ {3} b ^ {3} + b ^ {6})}$
$w=a(a^{6}+3a^{3}b^{3}+2b^{6})$ ${\ displaystyle w = a (a ^ {6} + 3a ^ {3} b ^ {3} + 2b ^ {6})}$

$x=a^{2}(b^{6}-7)+9ac-3c^{2}$ ${\ displaystyle x = a ^ {2} (b ^ {6} -7) + 9ac-3c ^ {2}}$
$y=a^{2}{\big [}b^{3}(2b^{3}+9)+7{\big ]}-3ac(2b^{3}+3)+3c^{2}$ ${\ displaystyle y = a ^ {2} {\ big [} b ^ {3} (2b ^ {3} +9) +7 {\ big]} - 3ac (2b ^ {3} +3) + 3c ^ {2}}$
$z=a^{2}b{\big [}b^{3}(b^{3}+3)+2{\big ]}-3abc(b^{3}+2)+3bc^{2}$ ${\ displaystyle z = a ^ {2} b {\ big [} b ^ {3} (b ^ {3} +3) +2 {\ big]} - 3abc (b ^ {3} +2) + 3bc ^ {2}}$
$w=a^{2}b{\big [}b^{3}(b^{3}+6)+11{\big ]}-3abc(b^{3}+4)+3bc^{2}$ ${\ displaystyle w = a ^ {2} b {\ big [} b ^ {3} (b ^ {3} +6) +11 {\ big]} - 3abc (b ^ {3} +4) + 3bc ^ {2}}$

$x=3a^{2}(b^{6}-7)-9ac-c^{2}$ ${\ displaystyle x = 3a ^ {2} (b ^ {6} -7) -9ac-c ^ {2}}$
$y=3a^{2}{\big [}b^{3}(2b^{3}-9)+7{\big ]}-3ac(2b^{3}-3)+c^{2}$ ${\ displaystyle y = 3a ^ {2} {\ big [} b ^ {3} (2b ^ {3} -9) +7 {\ big]} - 3ac (2b ^ {3} -3) + c ^ {2}}$
$z=3a^{2}b{\big [}b^{3}(b^{3}-6)+11{\big ]}-3abc(b^{3}-4)+bc^{2}$ ${\ displaystyle z = 3a ^ {2} b {\ big [} b ^ {3} (b ^ {3} -6) +11 {\ big]} - 3abc (b ^ {3} -4) + bc ^ {2}}$
$w=3a^{2}b{\big [}b^{3}(b^{3}-3)+2{\big ]}-3abc(b^{3}-2)+bc^{2}$ ${\ displaystyle w = 3a ^ {2} b {\ big [} b ^ {3} (b ^ {3} -3) +2 {\ big]} - 3abc (b ^ {3} -2) + bc ^ {2}}$

Roger Heath-Brown [1] , 1993

$x=9a^{4}$ ${\ displaystyle x = 9a ^ {4}}$
$y=3a-9a^{4}$ ${\ displaystyle y = 3a-9a ^ {4}}$
$z=1-9a^{3}$ ${\ displaystyle z = 1-9a ^ {3}}$
$w=1$ ${\ displaystyle w = 1}$

Mordell , 1956

$x=9a^{3}b+b^{4}$ ${\ displaystyle x = 9a ^ {3} b + b ^ {4}}$
$y=9a^{4}$ ${\ displaystyle y = 9a ^ {4}}$
$z=-b^{4}$ ${\ displaystyle z = -b ^ {4}}$
$w=9a^{4}+3ab^{3}$ ${\ displaystyle w = 9a ^ {4} + 3ab ^ {3}}$

$x=9a^{3}b-b^{4}$ ${\ displaystyle x = 9a ^ {3} bb ^ {4}}$
$y=9a^{4}-3ab^{3}$ ${\ displaystyle y = 9a ^ {4} -3ab ^ {3}}$
$z=b^{4}$ ${\ displaystyle z = b ^ {4}}$
$w=9a^{4}$ ${\ displaystyle w = 9a ^ {4}}$

$x=9a^{3}b+b^{4}$ ${\ displaystyle x = 9a ^ {3} b + b ^ {4}}$
$y=9a^{3}b-b^{4}$ ${\ displaystyle y = 9a ^ {3} bb ^ {4}}$
$z=9a^{4}-3ab^{3}$ ${\ displaystyle z = 9a ^ {4} -3ab ^ {3}}$
$w=9a^{4}+3ab^{3}$ ${\ displaystyle w = 9a ^ {4} + 3ab ^ {3}}$

The solution obtained by the method of algebraic geometry ( en: Fermat cubic )

$x=3a\left(a^{2}+ab+b^{2}\right)-9$ ${\ displaystyle x = 3a \ left (a ^ {2} + ab + b ^ {2} \ right) -9}$
$y=\left(a^{2}+ab+b^{2}\right)^{2}-9a$ ${\ displaystyle y = \ left (a ^ {2} + ab + b ^ {2} \ right) ^ {2} -9a}$
$z=3\left(a^{2}+ab+b^{2}\right)(a+b)+9$ ${\ displaystyle z = 3 \ left (a ^ {2} + ab + b ^ {2} \ right) (a + b) +9}$
$w=\left(a^{2}+ab+b^{2}\right)^{2}+9(a+b)$ ${\ displaystyle w = \ left (a ^ {2} + ab + b ^ {2} \ right) ^ {2} +9 (a + b)}$

Ramanujan

$x=3a^{2}+5ab-5b^{2}$ ${\ displaystyle x = 3a ^ {2} + 5ab-5b ^ {2}}$
$y=4a^{2}-4ab+6b^{2}$ ${\ displaystyle y = 4a ^ {2} -4ab + 6b ^ {2}}$
$z=5a^{2}-5ab-3b^{2}$ ${\ displaystyle z = 5a ^ {2} -5ab-3b ^ {2}}$
$w=6a^{2}-4ab+4b^{2}$ ${\ displaystyle w = 6a ^ {2} -4ab + 4b ^ {2}}$

$x=a^{7}-3a^{4}(1+b)+a(2+6b+3b^{2})$ ${\ displaystyle x = a ^ {7} -3a ^ {4} (1 + b) + a (2 + 6b + 3b ^ {2})}$
$y=2a^{6}-3a^{3}(1+2b)+1+3b+3b^{2}$ ${\ displaystyle y = 2a ^ {6} -3a ^ {3} (1 + 2b) + 1 + 3b + 3b ^ {2}}$
$z=a^{6}-1-3b-3b^{2}$ ${\ displaystyle z = a ^ {6} -1-3b-3b ^ {2}}$
$w=a^{7}-3a^{4}b+a(3b^{2}-1)$ ${\ displaystyle w = a ^ {7} -3a ^ {4} b + a (3b ^ {2} -1)}$

$x=-a^{2}+9ab+b^{2}$ ${\ displaystyle x = -a ^ {2} + 9ab + b ^ {2}}$
$y=a^{2}+7ab-9b^{2}$ ${\ displaystyle y = a ^ {2} + 7ab-9b ^ {2}}$
$z=2a^{2}-4ab+12b^{2}$ ${\ displaystyle z = 2a ^ {2} -4ab + 12b ^ {2}}$
$w=2a^{2}+10b^{2}$ ${\ displaystyle w = 2a ^ {2} + 10b ^ {2}}$

Unknown author, 1825

$x=a^{9}-3^{6}$ ${\ displaystyle x = a ^ {9} -3 ^ {6}}$
$y=-a^{9}+3^{5}a^{3}+3^{6}$ ${\ displaystyle y = -a ^ {9} + 3 ^ {5} a ^ {3} + 3 ^ {6}}$
$z=3^{3}a^{6}+3^{5}a^{3}$ ${\ displaystyle z = 3 ^ {3} a ^ {6} + 3 ^ {5} a ^ {3}}$
$w=3^{2}a^{7}+3^{4}a^{4}+3^{6}a$ ${\ displaystyle w = 3 ^ {2} a ^ {7} + 3 ^ {4} a ^ {4} + 3 ^ {6} a}$

D. Lemer, 1955

$x=3888a^{10}-135a^{4}$ ${\ displaystyle x = 3888a ^ {10} -135a ^ {4}}$
$y=-3888a^{10}-1296a^{7}-81a^{4}+3a$ ${\ displaystyle y = -3888a ^ {10} -1296a ^ {7} -81a ^ {4} + 3a}$
$z=3888a^{9}+648a^{6}-9a^{3}+1$ ${\ displaystyle z = 3888a ^ {9} + 648a ^ {6} -9a ^ {3} +1}$
$w=1$ ${\ displaystyle w = 1}$

V. B. Labkovsky

$x=4b^{2}-11b-21$ ${\ displaystyle x = 4b ^ {2} -11b-21}$
$y=3b^{2}+11b-28$ ${\ displaystyle y = 3b ^ {2} + 11b-28}$
$z=5b^{2}-7b+42$ ${\ displaystyle z = 5b ^ {2} -7b + 42}$
$w=6b^{2}-7b+35$ ${\ displaystyle w = 6b ^ {2} -7b + 35}$

Hardy and Wright

$x=a(a^{3}-2b^{3})$ ${\ displaystyle x = a (a ^ {3} -2b ^ {3})}$
$y=b(2a^{3}-b^{3})$ ${\ displaystyle y = b (2a ^ {3} -b ^ {3})}$
$z=b(a^{3}+b^{3})$ ${\ displaystyle z = b (a ^ {3} + b ^ {3})}$
$w=a(a^{3}+b^{3})$ ${\ displaystyle w = a (a ^ {3} + b ^ {3})}$

$x=a(a^{3}-b^{3})$ ${\ displaystyle x = a (a ^ {3} -b ^ {3})}$
$y=b(a^{3}-b^{3})$ ${\ displaystyle y = b (a ^ {3} -b ^ {3})}$
$z=b(2a^{3}+b^{3})$ ${\ displaystyle z = b (2a ^ {3} + b ^ {3})}$
$w=a(a^{3}+2b^{3})$ ${\ displaystyle w = a (a ^ {3} + 2b ^ {3})}$

G. Alexandrov, 1972

$x=7a^{2}+17ab-6b^{2}$ ${\ displaystyle x = 7a ^ {2} + 17ab-6b ^ {2}}$
$y=42a^{2}-17ab-b^{2}$ ${\ displaystyle y = 42a ^ {2} -17ab-b ^ {2}}$
$z=56a^{2}-35ab+9b^{2}$ ${\ displaystyle z = 56a ^ {2} -35ab + 9b ^ {2}}$
$w=63a^{2}-35ab+8b^{2}$ ${\ displaystyle w = 63a ^ {2} -35ab + 8b ^ {2}}$

$x=7a^{2}+17ab-17b^{2}$ ${\ displaystyle x = 7a ^ {2} + 17ab-17b ^ {2}}$
$y=17a^{2}-17ab-7b^{2}$ ${\ displaystyle y = 17a ^ {2} -17ab-7b ^ {2}}$
$z=14a^{2}-20ab+20b^{2}$ ${\ displaystyle z = 14a ^ {2} -20ab + 20b ^ {2}}$
$w=20a^{2}-20ab+14b^{2}$ ${\ displaystyle w = 20a ^ {2} -20ab + 14b ^ {2}}$

$x=21a^{2}+23ab-19b^{2}$ ${\ displaystyle x = 21a ^ {2} + 23ab-19b ^ {2}}$
$y=19a^{2}-23ab-21b^{2}$ ${\ displaystyle y = 19a ^ {2} -23ab-21b ^ {2}}$
$z=18a^{2}+4ab+28b^{2}$ ${\ displaystyle z = 18a ^ {2} + 4ab + 28b ^ {2}}$
$w=28a^{2}+4ab+18b^{2}$ ${\ displaystyle w = 28a ^ {2} + 4ab + 18b ^ {2}}$

$x=3a^{2}+41ab-37b^{2}$ ${\ displaystyle x = 3a ^ {2} + 41ab-37b ^ {2}}$
$y=37a^{2}-41ab-3b^{2}$ ${\ displaystyle y = 37a ^ {2} -41ab-3b ^ {2}}$
$z=36a^{2}-68ab+46b^{2}$ ${\ displaystyle z = 36a ^ {2} -68ab + 46b ^ {2}}$
$w=46a^{2}-68ab+36b^{2}$ ${\ displaystyle w = 46a ^ {2} -68ab + 36b ^ {2}}$

$x=-4a^{2}+22ab-9b^{2}$ ${\ displaystyle x = -4a ^ {2} + 22ab-9b ^ {2}}$
$y=36a^{2}-22ab+b^{2}$ ${\ displaystyle y = 36a ^ {2} -22ab + b ^ {2}}$
$z=40a^{2}-40ab+12b^{2}$ ${\ displaystyle z = 40a ^ {2} -40ab + 12b ^ {2}}$
$w=48a^{2}-40ab+10b^{2}$ ${\ displaystyle w = 48a ^ {2} -40ab + 10b ^ {2}}$

Ajai Choudhry, 1998 ^[4]

$dx_{1}=(a^{4}+2a^{3}b+3a^{2}b^{2}+2ab^{3}+b^{4})+(2a+b)c^{3},$ ${\ displaystyle dx_ {1} = (a ^ {4} + 2a ^ {3} b + 3a ^ {2} b ^ {2} + 2ab ^ {3} + b ^ {4}) + (2a + b ) c ^ {3},}$
$dx_{2}=-\{a^{4}+2a^{3}b+3a^{2}b^{2}+2ab^{3}+b^{4}-(a-b)c^{3}\},$ ${\ displaystyle dx_ {2} = - \ {a ^ {4} + 2a ^ {3} b + 3a ^ {2} b ^ {2} + 2ab ^ {3} + b ^ {4} - (ab) c ^ {3} \},}$
$dx_{3}=c(-a^{3}+b^{3}+c^{3}),$ ${\ displaystyle dx_ {3} = c (-a ^ {3} + b ^ {3} + c ^ {3}),}$
$dx_{4}=-\{(2a^{3}+3a^{2}b+3ab^{2}+b^{3})c+c^{4}\},$ ${\ displaystyle dx_ {4} = - \ {(2a ^ {3} + 3a ^ {2} b + 3ab ^ {2} + b ^ {3}) c + c ^ {4} \},}$

where are the numbers $a, b, c$ ${\ displaystyle a, b, c}$ Are arbitrary integers, and the number $d\neq 0$ ${\ displaystyle d \ neq 0}$ selected so that the condition is satisfied $(x_{1},x_{2},x_{3},x_{4})=1$ ${\ displaystyle (x_ {1}, x_ {2}, x_ {3}, x_ {4}) = 1}$ .

Koroviev, 2012

$x=-(2a^{2}-2ab-b^{2})cd^{3}-(a^{2}-ab+b^{2})^{2}c^{4}$ ${\ displaystyle x = - (2a ^ {2} -2ab-b ^ {2}) cd ^ {3} - (a ^ {2} -ab + b ^ {2}) ^ {2} c ^ {4 }}$
$y=\quad (2a^{2}-2ab-b^{2})c^{3}d+(a^{2}-ab+b^{2})^{2}d^{4}$ ${\ displaystyle y = \ quad (2a ^ {2} -2ab-b ^ {2}) c ^ {3} d + (a ^ {2} -ab + b ^ {2}) ^ {2} d ^ { four}}$
$z=\quad (a^{2}+2ab-2b^{2})c^{3}d-(a^{2}-ab+b^{2})^{2}d^{4}$ ${\ displaystyle z = \ quad (a ^ {2} + 2ab-2b ^ {2}) c ^ {3} d- (a ^ {2} -ab + b ^ {2}) ^ {2} d ^ {four}}$
$w=\quad (a^{2}+2ab-2b^{2})cd^{3}-(a^{2}-ab+b^{2})^{2}c^{4}$ ${\ displaystyle w = \ quad (a ^ {2} + 2ab-2b ^ {2}) cd ^ {3} - (a ^ {2} -ab + b ^ {2}) ^ {2} c ^ { four}}$

Where $a$ ${\ displaystyle a}$ , $b\,,\,c$ ${\ displaystyle b \ ,, \, c}$ and $d$ ${\ displaystyle d}$ - any integers. ^[five]

Notes

↑ ¹ ² Cohen, Henri. 6.4 Diophantine Equations of Degree 3 // Number Theory - Volume I: Tools and Diophantine Equations. - Springer-Verlag , 2007 .-- Vol. 239. - ISBN 978-0-387-49922-2 .
↑ An introduction to the theory of numbers. - First ed. - Clarendon Press, 1938.
↑ Quote from the section "1.3.7 Equation $x^{3}+y^{3}+z^{3}=t^{3}$ ${\ displaystyle x ^ {3} + y ^ {3} + z ^ {3} = t ^ {3}}$ "from Hardy and Wright
↑ Ajai Choudhry. On Equal Sums of Cubes . Rocky Mountain J. Math. Volume 28, Number 4 (1998), 1251-1257.
↑ In many cases, numbers $x, y, z, w$ ${\ displaystyle x, y, z, w}$ have common divisors. To get a primitive four of numbers, it is enough to reduce each of the numbers by their greatest common divisor .

Literature

Hardy G. Twelve Lectures on Ramanujan. - M .: Institute for Computer Research, 2002. - 336 p.
V. Sierpinsky . §15. Solution of equations in rational numbers // On the solution of equations in integers . - M .: Fizmatlit, 1961 .-- 88 p.
E. Rowland. Known families of integer solutions to $x^{3}+y^{3}+z^{3}=n$ ${\ displaystyle x ^ {3} + y ^ {3} + z ^ {3} = n}$ (English) : journal. Archived on September 27, 2013.
Labkovsky's decision (Task No. 2)
Sizyi S. V. 20. Comparisons of any degree in a simple module // Lectures on number theory: a manual for mathematical specialties . - Yekaterinburg: Ural State University named after A.M. Gorky , 1999.
Weisstein, Eric W. Diophantine Equation — 3rd Powers on Wolfram MathWorld .

[cohen1-1] ¹ ² Cohen, Henri. 6.4 Diophantine Equations of Degree 3 // Number Theory - Volume I: Tools and Diophantine Equations. - Springer-Verlag , 2007 .-- Vol. 239. - ISBN 978-0-387-49922-2 .

[2] An introduction to the theory of numbers. - First ed. - Clarendon Press, 1938.

[3] Quote from the section "1.3.7 Equation $x^{3}+y^{3}+z^{3}=t^{3}$ ${\ displaystyle x ^ {3} + y ^ {3} + z ^ {3} = t ^ {3}}$ "from Hardy and Wright

[4] Ajai Choudhry. On Equal Sums of Cubes . Rocky Mountain J. Math. Volume 28, Number 4 (1998), 1251-1257.

[5] In many cases, numbers $x, y, z, w$ ${\ displaystyle x, y, z, w}$ have common divisors. To get a primitive four of numbers, it is enough to reduce each of the numbers by their greatest common divisor .