Question

Разложение на множители. 1) Решить в целых числах уравнение x^3 - y^3 = 2044 Выделение целой части. 2) Найти натуральные решения уравнения x·y - 11·(x + y) = 1

Answer 1

1) x³-y³=(x-y)(x²+xy+y²)

(x-y)(x²+xy+y²)=2044

(x-y)(x²+xy+y²)=2²·7·73

$\left \{ {{x-y=1} \atop {x^2+xy+y^2=2044}} \right.$

$\left \{ {{x-y=2} \atop {x^2+xy+y^2=1022}} \right. \left \{ {{x-y=4} \atop {x^2+xy+y^2=511}} \right. \left \{ {{x-y=7} \atop {x^2+xy+y^2=292}} \right. \left \{ {{x-y=14} \atop {x^2+xy+y^2=146}} \right. \left \{ {{x-y=28} \atop {x^2+xy+y^2=73}} \right.$

$\left \{ {{x-y=73} \atop {x^2+xy+y^2=28}} \right. \left \{ {{x-y=146} \atop {x^2+xy+y^2=14}} \right. \left \{ {{x-y=292} \atop {x^2+xy+y^2=7}} \right. \left \{ {{x-y=511} \atop {x^2+xy+y^2=4}} \right.\left \{ {{x-y=1022} \atop {x^2+xy+y^2=2}} \right.$

$\left \{ {{x-y=2044} \atop {x^2+xy+y^2=1}} \right.$

и

$\left \{ {{x-y=-1} \atop {x^2+xy+y^2=-2044}} \right.$

$\left \{ {{x-y=-2} \atop {x^2+xy+y^2=-1022}} \right. \left \{ {{x-y=-4} \atop {x^2+xy+y^2=-511}} \right. \left \{ {{x-y=-7} \atop {x^2+xy+y^2=-292}} \right. \left \{ {{x-y=-14} \atop {x^2+xy+y^2=-146}} \right. \left \{ {{x-y=-28} \atop {x^2+xy+y^2=-73}} \right.$

$\left \{ {{x-y=-73} \atop {x^2+xy+y^2=-28}} \right. \left \{ {{x-y=-146} \atop {x^2+xy+y^2=-14}} \right. \left \{ {{x-y=-292} \atop {x^2+xy+y^2=-7}} \right. \left \{ {{x-y=-511} \atop {x^2+xy+y^2=-4}} \right.\left \{ {{x-y=-1022} \atop {x^2+xy+y^2=-2}} \right.$

$\left \{ {{x-y=-2044} \atop {x^2+xy+y^2=-1}} \right.$

Решим эти 24 системы подстановки и получим решение системы:

{x-y=4

{x^2+xy+y^2=511

ответ (15;11)

2) xy-11x-11y=1⇒  x(y-11)=1+11y

$x=\frac{11y+1}{y-11}=11+\frac{122}{y-11}$

y-11  делится на 122 нацело

при y=12;тогда х=133;  

при y=13;тогда х=72;

при y=72;  тогда x=13

при y=133; тогда х=12