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October 21, 2019

Rational form - RMO 2019 Problem 1 Solution

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Suppose x is a non zero real number such that both \( x^5 \) and \( 20 x + \frac{19}{x} \) are rational numbers. Prove that x is a rational number. 

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Regional Math Olympiad, 2019 Problem 1[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]3/10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" open="off"]Challenges and Thrills in Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]Notice that you can represent higher powers by smaller powers.  Suppose \( 20x + \frac{19}{x} = r \) where r is rational.  Multiply both sides by x to get \( 20 x^2 = r \cdot x - 19 \) or \( x^2 = r_1 x + r_2 \) (that is divide by 20 to have r/20 as the coefficient of x and -19/20 as the x free number. both of these are rationals. therefore we name them \( r_1 \) and \(r_2 \) )

Now use the fact that \( x^5 \) is rational. 

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Suppose \( x^5 = r_3 \) (some rational number).  Then \( x^2 \cdot x^2 \cdot x = r_3 \)  But we know \( x^2 = r_1 x + r_2 \)  Replacing we have \( (r_1 \cdot x + r_2 )^2 \cdot x = r_3\)  Expanding we have \( r_1^2 x^3 + 2r_1r_2 \cdot x^2 + r_2^2 x = r_3 \) Again we will replace \( x^2 \) to have  \( r_1^2 \cdot (r_1 x + r_2) x + 2r_1r_2 \cdot (r_1 x + r_2) + r_2^2 x = r_3 \)  

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Expand \( r_1^2 \cdot (r_1 x + r_2) x + 2r_1r_2 \cdot (r_1 x + r_2) + r_2^2 x = r_3 \) \( r_1^3 x^2 + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3 \) Making one final replacement of x^2 and noting that squaring, adding, multiplying rationals gives rationals we have  \( r_1^3 (r_1 x + r_2) + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3 \) \( r_1^4 x + r_1^3r_2 + r_1^2 r_2 x + 2r_1^2 r_2 x + r_2^2 x + 2 r_1 r_2^2 = r_3 \) or we have x as a ratio of rationals. 

Hence x is rational.

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