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QUestion

how to solve these type of questions ? i have tried logarithm and inequalities but could not pin point the exact and correct method.

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Such problems are based on relation of the surd with its conjugate

if $(2+\sqrt3)^k=a+b\sqrt3$

then $(2-\sqrt3)^k=a-b\sqrt3$

Where a,b, are integers. So it follows that

$(2+\sqrt3)^k+(2-\sqrt3)^k=2a $ is integer.

Now note that $(2-\sqrt3)^k<1$ implying $1-(2-\sqrt3)^k$ is fractional part of $(2+\sqrt3)^k$. So $1-f=(2-\sqrt3)^k$ and $x(1-f)=(2+\sqrt3)^k(2-\sqrt3)^k=1^k=1$

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  • $\begingroup$ and "fractional part of $(2+\sqrt{3})^k$." $\endgroup$ – OR. Dec 20 '13 at 16:06
  • $\begingroup$ @zimbra314, please rectify the last step $1^k=1$, +1 though $\endgroup$ – lab bhattacharjee Dec 20 '13 at 16:07
  • $\begingroup$ "Now note that (2−3√)k<1 implying 1−(2−3√)k is fractional part of (2+3√)k " can u further explain this part $\endgroup$ – Amit Dec 20 '13 at 16:12
  • $\begingroup$ @Amit, together with the line before, it follows that $2a$ is the smallest integer greater than $(2+\sqrt3)^k$, and hence the greatest integer smaller than $(2+\sqrt3)^k$ is $2a-1$. So the fractional part is $(2+\sqrt3)^k-(2a-1)$. $\endgroup$ – Carsten S Dec 20 '13 at 16:23
  • $\begingroup$ ok ..thanx that was helpful $\endgroup$ – Amit Dec 20 '13 at 16:30

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