# Problem of the week #102 - March 10th, 2014

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#### anemone

##### MHB POTW Director
Staff member
Represent the number $\sqrt{1342\sqrt{167}+2005}$ in the form where it contains only addition, subtraction, multiplication, division and square roots.

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#### anemone

##### MHB POTW Director
Staff member
Re: Problem of the week #102 -March 10th, 2014

Congratulations to the following members for their correct solutions: 1. MarkFL
3. soroban
4. mente oscura

Because 167 is prime so the cube root shall be of the form $a + b\sqrt{167}$

Cube both sides and equate to cube of RHS to get

$a^3 + 3a^2b\sqrt {167} + 3a * 167 b^2 + 167 b^3 \sqrt {167} = 2005 + 1342 \sqrt {167}$

Equate the rational parts on both sides to get

$a^3 + 501 ab^2 = 2005$ ...(1)

Equate their rational parts on both sides to get

$3a^2 b + 167 b^3 = 1342$ ... (1)

First check for positive integers and we see that a = 1 and b =2 satisfy (1) and (2) and hence solution is

$1 + 2 \sqrt {167}$

Solution from mente oscura:

$$\sqrt{1342 \sqrt{167}+2005}$$

$$\sqrt{1342 \sqrt{167}+2005}=\sqrt{(a+b)^3}$$

$$a^3+3a^2b+3ab^2+b^3=1342 \sqrt{167}+2005$$

$$Let \ a=x \sqrt{167}$$

Therefore:

$$a^3+3ab^2=1342 \sqrt{167} \rightarrow{} 167x^3 \cancel{\sqrt{167}}+3x \cancel{\sqrt{167}} \ b^2=1342 \cancel{\sqrt{167}}$$

$$b=\sqrt{\dfrac{1342-167x^3}{3x}}$$

$$For \ x=2 \rightarrow{}b=1 \ and \ a=2 \sqrt{167}$$

Same:

$$3x^2 \ 167b+b^3=2005$$

$$For \ b=1 \rightarrow{} x=2$$

Solution:
$$\sqrt{1342 \sqrt{167}+2005}=2 \sqrt{167}+1$$

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