Question-173162

Question Number 173162 by dragan91 last updated on 07/Jul/22

Answered by aleks041103 last updated on 07/Jul/22

x + y > 0

\Rightarrow x^{z} > y^{z}

z \in N \Rightarrow x > y

l e t x = y + t, t \in N

{(y + t)}^{z} - y^{z} = 2 y + t

z = 1 \Rightarrow (y + t) - y = t = 2 y + t \Rightarrow y = 0 \notin N

\Rightarrow z > 1

{(y + t)}^{z} - y^{z} = y^{z} ({(1 + \frac{t}{y})}^{z} - 1^{z}) =

= y^{z} \frac{t}{y} (1 + (1 + \frac{t}{y}) + {(1 + \frac{t}{y})}^{2} + \dots + {(1 + \frac{t}{y})}^{z - 1}) =

= {t y}^{z - 1} (1 + (1 + \frac{t}{y}) + {(1 + \frac{t}{y})}^{2} + \dots + {(1 + \frac{t}{y})}^{z - 1})

\Rightarrow {(y + t)}^{z} - y^{z} > {t y}^{z - 1} (2 + \frac{t}{y}) = 2 {t y}^{z - 1} + t^{2} y^{z - 2}

z ⩾ 2 \Rightarrow y^{z - 2} ⩾ 1, y^{z - 1} ⩾ y

\Rightarrow x^{z} - y^{z} ⩾ 2 t y + t^{2} = t (2 y + t) = t (x + y)

\Rightarrow \frac{x^{z} - y^{z}}{x + y} ⩾ t

i f \exists x, y, z \in N, s . t . x^{z} - y^{z} = x + y, t h e n

1 ⩾ t \in N \Rightarrow t = 1

\Rightarrow x = y + 1

Commented by dragan91 last updated on 08/Jul/22

well done sir

Commented by Tawa11 last updated on 11/Jul/22

Great sir

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