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Вопрос по алгебре:
Докажите что уравнение x^2-y^2=30 не имеет решения в целых числах.
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Автор:
amyalvarez
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Ответ:
x^2-y^2=30
(x-y)(x+y)=30
так как x,y- целые, то x-y и x+yтоже целые
30=1*30=(-1)*(-30)=2*15=(-2)*(-15)=(-3)*(-10)=3*10
решив 12 систем
x-y=1
x+y=30
x-y=30
x+y=1
x-y=-1
x+y=-30
x-y=-30
x+y=-1
и т.д. легко убедиться что целых решений на одна их систем не имеет, а значит и данное уравнение не разрешимо в целых числах, т.е. не имеет решения в целых числах
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