數學知識交流(6)

1 Let x = [x] + d where 0 < x < 1

[x^2] - [x]^2 = 2009

([x] + d)^2 - [x]^2 = 2009

d^2 + 2[x]d - 2009 = 0

d = -[x] + √([x^2] + 2009)

Since 0 < d < 1, we have

0 < √([x^2] + 2009) - [x] < 1

Solving it, [x] > 1004

So, we can construct infinity numbers such that [x^2] - [x]^2 = 2009 with respect to different unteger part (> 1004)

Foe example, if [x] = 5000, then d = -[x] + √([x^2] + 2009)
= 0.20089596

2 Let y = [y] + d where 0 < x < 1

[y^2] - [y]^2 = 11111

([y] + d)^2 - [y]^2 = 11111

y^2 + 2[y]d - 11111 = 0

d = -[y] + √([y^2] + 11111)

Since 0 < d < 1, we have

0 < √([y^2] + 11111) - [y] < 1

Solving it, [y] > 5555

So, we can construct infinity numbers such that [y^2] - [y]^2 = 11111 with respect to different unteger part (> 5555)

Foe example, if [y] = 5556, then d = -[y] + √([y^2] + 11111)
= 0.99982